Left Rotation of a String
Last Updated :
11 Nov, 2024
Given a string s and an integer d, the task is to left rotate the string by d positions.
Examples:
Input: s = "GeeksforGeeks", d = 2
Output: "eksforGeeksGe"
Explanation: After the first rotation, string s becomes "eeksforGeeksG" and after the second rotation, it becomes "eksforGeeksGe".
Input: s = "qwertyu", d = 2
Output: "ertyuqw"
Explanation: After the first rotation, string s becomes "wertyuq" and after the second rotation, it becomes "ertyuqw".
[Naive Approach] Left Rotate one by one
The idea is to store the first character in a variable and shift all the remaining characters to the left by one position, then place the first character at the end of string. This process is repeated d times.
C++
// C++ Program to left rotate the string by d
// positions by rotating one element at a time
#include <iostream>
#include <string>
using namespace std;
void rotateString(string &s, int d) {
int n = s.size();
// Repeat the rotation d times
for (int i = 0; i < d; i++) {
// Left rotate the string by one position
int first = s[0];
for (int j = 0; j < n - 1; j++)
s[j] = s[j + 1];
// Place the first character at the end
s[n - 1] = first;
}
}
int main() {
string s = "GeeksforGeeks";
int d = 2;
rotateString(s, d);
cout << s << endl;
return 0;
}
// C Program to left rotate the string by d positions
// by rotating one element at a time
#include <stdio.h>
#include <string.h>
void rotateString(char s[], int d) {
int n = strlen(s);
// Repeat the rotation d times
for (int i = 0; i < d; i++) {
// Left rotate the string by one position
char first = s[0];
for (int j = 0; j < n - 1; j++)
s[j] = s[j + 1];
// Place the first character at the end
s[n - 1] = first;
}
}
int main() {
char s[] = "GeeksforGeeks";
int d = 2;
rotateString(s, d);
printf("%s\n", s);
return 0;
}
// Java Program to left rotate the string by d positions
// by rotating one element at a time
import java.util.Arrays;
class GfG {
static String rotateString(String s, int d) {
// Convert the string to a char array
char[] charArray = s.toCharArray();
int n = charArray.length;
// Perform the rotation d times
for (int i = 0; i < d; i++) {
// Store the first character
char first = charArray[0];
// Shift each character one position to
// the left
for (int j = 0; j < n - 1; j++)
charArray[j] = charArray[j + 1];
// Move the first character to the end
charArray[n - 1] = first;
}
return new String(charArray);
}
public static void main(String[] args) {
String s = "GeeksforGeeks";
int d = 2;
String rotatedString = rotateString(s, d);
System.out.println(rotatedString);
}
}
# python Program to left rotate the string by d
# positions by rotating one element at a time
def rotateString(s, d):
# Convert the string to a list of
# characters
s = list(s)
n = len(s)
# Perform the rotation d times
for _ in range(d):
# Store the first character
first = s[0]
# Shift each character one
# position to the left
for i in range(n - 1):
s[i] = s[i + 1]
# Move the first character to the end
s[n - 1] = first
# Convert the list back to a string
return ''.join(s)
s = "GeeksforGeeks"
d = 2
rotatedString = rotateString(s, d)
print(rotatedString)
// C# Program to left rotate the string by d positions
// by rotating one element at a time
using System;
class GfG {
static string rotateString(string s, int d) {
// Convert the string to a character array
char[] charArray = s.ToCharArray();
int n = charArray.Length;
// Perform the rotation d times
for (int i = 0; i < d; i++) {
// Store the first character
char first = charArray[0];
// Shift each character one position to
// the left
for (int j = 0; j < n - 1; j++)
charArray[j] = charArray[j + 1];
// Move the first character to the end
charArray[n - 1] = first;
}
// Convert the character array to a string
return new string(charArray);
}
static void Main() {
string s = "GeeksforGeeks";
int d = 2;
string rotatedString = rotateString(s, d);
Console.WriteLine(rotatedString);
}
}
// Javascript Program to left rotate the string by d positions
// by rotating one element at a time
function rotateString(s, d) {
// Convert the string to an array
let charArray = s.split('');
let n = charArray.length;
// Perform the rotation d times
for (let i = 0; i < d; i++) {
// Store the first character
let first = charArray[0];
// Shift each character one position to the left
for (let j = 0; j < n - 1; j++)
charArray[j] = charArray[j + 1];
// Move the first character to the end
charArray[n - 1] = first;
}
// Convert the array back to a string
return charArray.join('');
}
let s = "GeeksforGeeks";
let d = 2;
let rotatedString = rotateString(s, d);
console.log(rotatedString);
Time Complexity: O(n*d), the outer loop runs d times, and inner loop runs n times.
Auxiliary Space: O(1) if the string is mutable, like in C++. For immutable strings like in Java, C#, Python and Javascript an extra character array of size n is used, so the space complexity will be O(n).
[Better Approach] Using Temporary Char Array
The idea is to use a temporary character array of size n (size of original string). If we left rotate the string by d positions, the last n – d elements will be at the front and the first d elements will be at the end.
- Copy the last (n – d) elements of original string into the first n – d positions of temporary array.
- Then, copy the first d elements of the original string to the end of temporary array.
- Finally, convert the temporary char array to the string.
C++
// C++ program to left rotate a string by d
// position using a temporary array
#include <iostream>
#include <string>
using namespace std;
string rotateString(string &s, int d) {
int n = s.length();
// Handle cases where d > n
d = d % n;
char temp[n];
// Copy the last (n - d) characters
// to the start of temp Array
for (int i = 0; i < n - d; i++)
temp[i] = s[d + i];
// Copy the first d characters to the end
// of temp Array
for (int i = 0; i < d; i++)
temp[n - d + i] = s[i];
// Convert temp array to the string
return string(temp, n);
}
int main() {
string s = "GeeksforGeeks";
int d = 2;
string rotatedString = rotateString(s, d);
cout << rotatedString << endl;
return 0;
}
// Java program to left rotate a string by d position
// using a temporary array
import java.io.*;
class GfG {
static String rotateString(String s, int d) {
int n = s.length();
// Handle cases where d > n
d = d % n;
char[] temp = new char[n];
// Copy the last (n - d) characters to the
// start of temp array
for (int i = 0; i < n - d; i++)
temp[i] = s.charAt(d + i);
// Copy the first d characters to the end of
// temp array
for (int i = 0; i < d; i++)
temp[n - d + i] = s.charAt(i);
// Convert the temp array back to the String
return new String(temp);
}
public static void main(String[] args) {
String s = "GeeksforGeeks";
int d = 2;
String rotatedString = rotateString(s, d);
System.out.println(rotatedString);
}
}
# Python program to left rotate a string
# by d position using a temporary array
def rotateString(s, d):
n = len(s)
# Handle cases where d > n
d = d % n
# Create a temporary array of the
# same length as s
temp = [''] * n
# Copy the last (n - d) characters
# to the start of temp array
for i in range(n - d):
temp[i] = s[d + i]
# Copy the first d characters to the
#end of temp array
for i in range(d):
temp[n - d + i] = s[i]
# Convert temp array back to the string
return ''.join(temp)
s = "GeeksforGeeks"
d = 2
rotatedString = rotateString(s, d)
print(rotatedString)
// C# program to left rotate a string by d position
// using temporary array
using System;
class GfG {
static string rotateString(string s, int d) {
int n = s.Length;
// Handle cases where d > n
d = d % n;
char[] temp = new char[n];
// Copy the last (n - d) characters
// to the start of temp array
for (int i = 0; i < n - d; i++)
temp[i] = s[d + i];
// Copy the first d characters to the end
// of temp array
for (int i = 0; i < d; i++)
temp[n - d + i] = s[i];
// Convert temp array back to the string
return new string(temp);
}
static void Main() {
string s = "GeeksforGeeks";
int d = 2;
string rotatedString = rotateString(s, d);
Console.WriteLine(rotatedString);
}
}
// Javascript program to left rotate a string
// by d position using temporary array
function rotateString(s, d) {
let n = s.length;
// Handle cases where d > n
d = d % n;
let temp = new Array(n);
// Copy the last (n - d) characters to
// the start of temp array
for (let i = 0; i < n - d; i++)
temp[i] = s[d + i];
// Copy the first d characters
// to the end of temp array
for (let i = 0; i < d; i++)
temp[n - d + i] = s[i];
// Convert the array back to the string
return temp.join("");
}
let s = "GeeksforGeeks";
let d = 2;
let rotatedString = rotateString(s, d);
console.log(rotatedString);
Time Complexity: O(n), as we are visiting each element only twice.
Auxiliary Space: O(n), as we are using an additional character array.
[Expected Approach - 1] Using Juggling Algorithm
The idea behind the juggling algorithm is that we can rotate all the elements in cycle. Each cycle is independent and represents a group of elements that will shift among themselves during the rotation. If the starting index of a cycle is i, then next elements of the cycle will be present at indices (i + d) % n, (i + 2d) % n, (i + 3d) % n ... and so on till we reach back to index i. The total number of cycles will be GCD of n and d. And, we perform a single left rotation within each cycle.
To know more about the Juggling algorithm, refer this article - Juggling Algorithm for Array Rotation.
C++
// C++ Program to left rotate the string by d positions
// using Juggling Algorithm
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
void rotateString(string &s, int d) {
int n = s.size();
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
int cycles = __gcd(n, d);
// Perform a left rotation within each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
char startChar = s[i];
// Start index of current cycle
int currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while (true) {
nextIdx = (currIdx + d) % n;
if (nextIdx == i)
break;
// Update the next index with the current element
s[currIdx] = s[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle
// at the last index of the cycle
s[currIdx] = startChar;
}
}
int main() {
string s = "GeeksforGeeks";
int d = 2;
rotateString(s, d);
cout << s << endl;
return 0;
}
// C Program to left rotate the string by d positions
// using Juggling Algorithm
#include <stdio.h>
#include <string.h>
void rotateString(char s[], int d) {
int n = strlen(s);
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the
// rotation
int cycles = gcd(n, d);
// Perform a left rotation within each cycle
for (int i = 0; i < cycles; i++) {
// Start element of the current cycle
char startChar = s[i];
// Start index of the current cycle
int currIdx = i, nextIdx;
// Rotate elements until we return to the
// start of the cycle
while (1) {
nextIdx = (currIdx + d) % n;
if (nextIdx == i)
break;
// Update the current index with the
// element at the next index
s[currIdx] = s[nextIdx];
currIdx = nextIdx;
}
// Place the start element of the current
// cycle at the last index
s[currIdx] = startChar;
}
}
int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
int main() {
char s[] = "GeeksforGeeks";
int d = 2;
rotateString(s, d);
printf("%s\n", s);
return 0;
}
// Java Program to left rotate the string by d positions
// using Juggling Algorithm
import java.io.*;
class GfG {
static String rotateString(String s, int d) {
int n = s.length();
// Handle the case where
// d > size of the string
d %= n;
// Calculate the number of
// cycles (GCD of n and d)
int cycles = gcd(n, d);
// Convert string to character array
char[] arr = s.toCharArray();
// Perform a left rotation within each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
char temp = arr[i];
int j = i;
while (true) {
int k = (j + d) % n;
if (k == i) {
break;
}
// Move the element to the next index
arr[j] = arr[k];
j = k;
}
// Place the saved element in the
// last position of the cycle
arr[j] = temp;
}
// Convert the rotated character
// array back to a string
return new String(arr);
}
// function to calculate GCD of two numbers
static int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static void main(String[] args) {
String s = "GeeksforGeeks";
int d = 2;
String rotatedString = rotateString(s, d);
System.out.println(rotatedString);
}
}
# python Program to left rotate the string by
# d positions using Juggling Algorithm
def gcd(a, b):
while b:
a, b = b, a % b
return a
def rotateString(s, d):
n = len(s)
# Handle the case where d > size of
# the string
d %= n
# Calculate the number of cycles (GCD
# of n and d)
cycles = gcd(n, d)
# Convert string to a list of characters
arr = list(s)
# Perfrom a left rotation wihtin each cycle
for i in range(cycles):
# Start element of current cycle
temp = arr[i]
j = i
while True:
k = (j + d) % n
if k == i:
break
# Move the element to the next
# index
arr[j] = arr[k]
j = k
# Place the saved element in the last
# position of the cycle
arr[j] = temp
# Convert the list of characters back to
# a string and return
return ''.join(arr)
s = "GeeksforGeeks"
d = 2
rotatedString = rotateString(s, d)
print(rotatedString)
// C# Program to left rotate the string by d positions
// using Juggling Algorithm
using System;
class GfG {
static int Gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
static string rotateString(string s, int d) {
int n = s.Length;
// Handle the case where d > size of the string
d %= n;
// Calculate the number of cycles (GCD of n and d)
int cycles = Gcd(n, d);
// Convert string to a character array
char[] arr = s.ToCharArray();
// Perform a left rotation within each cycle
for (int i = 0; i < cycles; i++) {
// Start element of the current cycle
char temp = arr[i];
int j = i;
while (true) {
int k = (j + d) % n;
if (k == i)
break;
// Move the element to the next index
arr[j] = arr[k];
j = k;
}
// Place the saved element in the last position
// of the cycle
arr[j] = temp;
}
// Convert the character array back to a string
return new string(arr);
}
static void Main() {
string s = "GeeksforGeeks";
int d = 2;
string rotatedString = rotateString(s, d);
Console.WriteLine(rotatedString);
}
}
// JavaScript Program to left rotate the string by d
// positions using Juggling Algorithm
function gcd(a, b) {
while (b !== 0) {
let temp = b;
b = a % b;
a = temp;
}
return a;
}
function rotateString(s, d) {
let n = s.length;
// Handle the case where d > size of the string
d %= n;
// Calculate the number of cycles (GCD of n and d)
let cycles = gcd(n, d);
// Convert string to a character array
let arr = s.split('');
// Perform a left rotation within each cycle
for (let i = 0; i < cycles; i++) {
// Start element of the current cycle
let temp = arr[i];
let j = i;
while (true) {
let k = (j + d) % n;
if (k === i) {
break;
}
// Move the element to the next index
arr[j] = arr[k];
j = k;
}
// Place the first element in the last position
// of the cycle
arr[j] = temp;
}
// Convert the character array back to a string
return arr.join('');
}
let s = "GeeksforGeeks";
let d = 2;
let rotatedString = rotateString(s, d);
console.log(rotatedString);
Time Complexity: O(n)
Auxiliary Space: O(1) if the string is mutable, like in C++. For immutable strings like in Java, C#, Python and Javascript an extra character array of size n is used, so the space complexity will be O(n).
[Expected Approach - 2] Using Reversal Algorithm
The idea is based on the observation that if we left rotate the string by d positions, the last (n – d) elements will be at the front and the first d elements will be at the end.
- Reverse the substring containing the first d elements of the string.
- Reverse the substring containing the last (n – d) elements of the string.
- Finally, reverse all the elements of the string.
C++
// C++ program to left rotate a string by d position
// using Reversal Algorithm
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
void rotateString(string &s, int d) {
int n = s.size();
// Handle the case where d > size of array
d %= n;
// Reverse the first d elements
reverse(s.begin(), s.begin() + d);
// Reverse the remaining n-d elements
reverse(s.begin() + d, s.end());
// Reverse the entire string
reverse(s.begin(), s.end());
}
int main() {
string s = "GeeksforGeeks";
int d = 2;
rotateString(s, d);
cout << s << endl;
return 0;
}
// Java program to left rotate a string by d position
// using Reversal Algorithm
import java.io.*;
class GfG {
static String rotateString(String s, int d) {
int n = s.length();
// Handle the case where d > size of string
d %= n;
// Convert string to a character array
char[] temp = s.toCharArray();
// Reverse the first d elements
reverse(temp, 0, d - 1);
// Reverse the remaining n-d elements
reverse(temp, d, n - 1);
// Reverse the entire array
reverse(temp, 0, n - 1);
// Convert the array back to a string and return
return new String(temp);
}
static void reverse(char[] temp, int start, int end) {
while (start < end) {
char c = temp[start];
temp[start] = temp[end];
temp[end] = c;
start++;
end--;
}
}
public static void main(String[] args) {
String s = "GeeksforGeeks";
int d = 2;
String rotatedString = rotateString(s, d);
System.out.println(rotatedString);
}
}
# Python program to left rotate a string by d positons
# using Reversal Algorithm
def rotateString(s, d):
n = len(s)
# Handle cases where d > n
d %= n
# Convert the string to a list of characters
temp = list(s)
# Reverse the first d elements
reverse(temp, 0, d - 1)
# Reverse the remaining n - d elements
reverse(temp, d, n - 1)
# Reverse the entire array
reverse(temp, 0, n - 1)
# Convert the list back to a string and return
return ''.join(temp)
def reverse(temp, start, end):
while start < end:
temp[start], temp[end] = temp[end], temp[start]
start += 1
end -= 1
s = "GeeksforGeeks"
d = 2
rotatedString = rotateString(s, d)
print(rotatedString)
// C++ program to left rotate a string by d positions
// using Reversal Algorithm
using System;
class GfG {
static string RotateString(string s, int d) {
int n = s.Length;
// Handle cases where d > n
d %= n;
// Convert the string to a character array
char[] temp = s.ToCharArray();
// Reverse the first d elements
Reverse(temp, 0, d - 1);
// Reverse the remaining n - d elements
Reverse(temp, d, n - 1);
// Reverse the entire array
Reverse(temp, 0, n - 1);
// Convert the character array back to a string
return new string(temp);
}
static void Reverse(char[] temp, int start, int end) {
while (start < end) {
char c = temp[start];
temp[start] = temp[end];
temp[end] = c;
start++;
end--;
}
}
static void Main() {
string s = "GeeksforGeeks";
int d = 2;
string rotatedString = RotateString(s, d);
Console.WriteLine(rotatedString);
}
}
// C++ program to left rotate a string by d position
// using Reversal Algorithm
function rotateString(s, d) {
const n = s.length;
// Handle cases where d > n
d %= n;
// Convert the string to a character array
let temp = s.split("");
// Reverse the first d elements
reverse(temp, 0, d - 1);
// Reverse the remaining n - d elements
reverse(temp, d, n - 1);
// Reverse the entire array
reverse(temp, 0, n - 1);
// Convert the array back to a string
return temp.join("");
}
function reverse(temp, start, end) {
while (start < end) {
// Swap elements
[temp[start], temp[end]]
= [temp[end], temp[start]];
start++;
end--;
}
}
let s = "GeeksforGeeks";
let d = 2;
let rotatedString = rotateString(s, d);
console.log(rotatedString);
Time Complexity: O(n), where n is the size of the given string.
Auxiliary Space: O(1) if the string is mutable, like in C++. For immutable strings like in Java, C#, python and Javascript, an extra character array of size n is used, so the space complexity will be O(n).
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Basic operations on String
Searching For Characters and Substring in a String in JavaEfficient String manipulation is very important in Java programming especially when working with text-based data. In this article, we will explore essential methods like indexOf(), contains(), and startsWith() to search characters and substrings within strings in Java.Searching for a Character in a
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Reverse a String – Complete TutorialGiven a string s, the task is to reverse the string. Reversing a string means rearranging the characters such that the first character becomes the last, the second character becomes second last and so on.Examples:Input: s = "GeeksforGeeks"Output: "skeeGrofskeeG"Explanation : The first character G mo
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Left Rotation of a StringGiven a string s and an integer d, the task is to left rotate the string by d positions.Examples:Input: s = "GeeksforGeeks", d = 2Output: "eksforGeeksGe" Explanation: After the first rotation, string s becomes "eeksforGeeksG" and after the second rotation, it becomes "eksforGeeksGe".Input: s = "qwer
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Sort string of charactersGiven a string of lowercase characters from 'a' - 'z'. We need to write a program to print the characters of this string in sorted order.Examples: Input : "dcab" Output : "abcd"Input : "geeksforgeeks"Output : "eeeefggkkorss"Naive Approach - O(n Log n) TimeA simple approach is to use sorting algorith
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Frequency of Characters in Alphabetical OrderGiven a string s, the task is to print the frequency of each of the characters of s in alphabetical order.Example: Input: s = "aabccccddd" Output: a2b1c4d3 Since it is already in alphabetical order, the frequency of the characters is returned for each character. Input: s = "geeksforgeeks" Output: e4
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Swap characters in a StringGiven a String S of length N, two integers B and C, the task is to traverse characters starting from the beginning, swapping a character with the character after C places from it, i.e. swap characters at position i and (i + C)%N. Repeat this process B times, advancing one position at a time. Your ta
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C Program to Find the Length of a StringThe length of a string is the number of characters in it without including the null character (‘\0’). In this article, we will learn how to find the length of a string in C.The easiest way to find the string length is by using strlen() function from the C strings library. Let's take a look at an exa
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How to insert characters in a string at a certain position?Given a string str and an array of indices chars[] that describes the indices in the original string where the characters will be added. For this post, let the character to be inserted in star (*). Each star should be inserted before the character at the given index. Return the modified string after
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Check if two strings are same or notGiven two strings, the task is to check if these two strings are identical(same) or not. Consider case sensitivity.Examples:Input: s1 = "abc", s2 = "abc" Output: Yes Input: s1 = "", s2 = "" Output: Yes Input: s1 = "GeeksforGeeks", s2 = "Geeks" Output: No Approach - By Using (==) in C++/Python/C#, eq
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Concatenating Two Strings in CConcatenating two strings means appending one string at the end of another string. In this article, we will learn how to concatenate two strings in C.The most straightforward method to concatenate two strings is by using strcat() function. Let's take a look at an example:C#include <stdio.h> #i
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Remove all occurrences of a character in a stringGiven a string and a character, remove all the occurrences of the character in the string.Examples: Input : s = "geeksforgeeks" c = 'e'Output : s = "gksforgks"Input : s = "geeksforgeeks" c = 'g'Output : s = "eeksforeeks"Input : s = "geeksforgeeks" c = 'k'Output : s = "geesforgees"Using Built-In Meth
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Binary String
Check if all bits can be made same by single flipGiven a binary string, find if it is possible to make all its digits equal (either all 0's or all 1's) by flipping exactly one bit. Input: 101Output: YeExplanation: In 101, the 0 can be flipped to make it all 1Input: 11Output: NoExplanation: No matter whichever digit you flip, you will not get the d
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Number of flips to make binary string alternate | Set 1Given a binary string, that is it contains only 0s and 1s. We need to make this string a sequence of alternate characters by flipping some of the bits, our goal is to minimize the number of bits to be flipped. Examples : Input : str = “001†Output : 1 Minimum number of flips required = 1 We can flip
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Binary representation of next numberGiven a binary string that represents binary representation of positive number n, the task is to find the binary representation of n+1. The binary input may or may not fit in an integer, so we need to return a string.Examples: Input: s = "10011"Output: "10100"Explanation: Here n = (19)10 = (10011)2n
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Min flips of continuous characters to make all characters same in a stringGiven a string consisting only of 1's and 0's. In one flip we can change any continuous sequence of this string. Find this minimum number of flips so the string consist of same characters only.Examples: Input : 00011110001110Output : 2We need to convert 1's sequenceso string consist of all 0's.Input
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Generate all binary strings without consecutive 1'sGiven an integer n, the task is to generate all binary strings of size n without consecutive 1's.Examples: Input : n = 4Output : 0000 0001 0010 0100 0101 1000 1001 1010Input : n = 3Output : 000 001 010 100 101Approach:The idea is to generate all binary strings of length n without consecutive 1's usi
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Find i'th Index character in a binary string obtained after n iterationsGiven a decimal number m, convert it into a binary string and apply n iterations. In each iteration, 0 becomes "01" and 1 becomes "10". Find the (based on indexing) index character in the string after the nth iteration. Examples: Input : m = 5, n = 2, i = 3Output : 1Input : m = 3, n = 3, i = 6Output
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Substring and Subsequence
All substrings of a given StringGiven a string s, containing lowercase alphabetical characters. The task is to print all non-empty substrings of the given string.Examples : Input : s = "abc"Output : "a", "ab", "abc", "b", "bc", "c"Input : s = "ab"Output : "a", "ab", "b"Input : s = "a"Output : "a"[Expected Approach] - Using Iterati
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Print all subsequences of a stringGiven a string, we have to find out all its subsequences of it. A String is said to be a subsequence of another String, if it can be obtained by deleting 0 or more character without changing its order.Examples: Input : abOutput : "", "a", "b", "ab"Input : abcOutput : "", "a", "b", "c", "ab", "ac", "
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Count Distinct SubsequencesGiven a string str of length n, your task is to find the count of distinct subsequences of it.Examples: Input: str = "gfg"Output: 7Explanation: The seven distinct subsequences are "", "g", "f", "gf", "fg", "gg" and "gfg" Input: str = "ggg"Output: 4Explanation: The four distinct subsequences are "",
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Count distinct occurrences as a subsequenceGiven two strings pat and txt, where pat is always shorter than txt, count the distinct occurrences of pat as a subsequence in txt.Examples: Input: txt = abba, pat = abaOutput: 2Explanation: pat appears in txt as below three subsequences.[abba], [abba]Input: txt = banana, pat = banOutput: 3Explanati
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Longest Common Subsequence (LCS)Given two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0. A subsequence is a string generated from the original string by deleting 0 or more characters, without changing the relative order of the remaining characters.
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Shortest Superstring ProblemGiven a set of n strings arr[], find the smallest string that contains each string in the given set as substring. We may assume that no string in arr[] is substring of another string.Examples: Input: arr[] = {"geeks", "quiz", "for"}Output: geeksquizforExplanation: "geeksquizfor" contains all the thr
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Printing Shortest Common SupersequenceGiven two strings s1 and s2, find the shortest string which has both s1 and s2 as its sub-sequences. If multiple shortest super-sequence exists, print any one of them.Examples:Input: s1 = "geek", s2 = "eke"Output: geekeExplanation: String "geeke" has both string "geek" and "eke" as subsequences.Inpu
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Shortest Common SupersequenceGiven two strings s1 and s2, the task is to find the length of the shortest string that has both s1 and s2 as subsequences.Examples: Input: s1 = "geek", s2 = "eke"Output: 5Explanation: String "geeke" has both string "geek" and "eke" as subsequences.Input: s1 = "AGGTAB", s2 = "GXTXAYB"Output: 9Explan
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Longest Repeating SubsequenceGiven a string s, the task is to find the length of the longest repeating subsequence, such that the two subsequences don't have the same string character at the same position, i.e. any ith character in the two subsequences shouldn't have the same index in the original string. Examples:Input: s= "ab
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Longest Palindromic Subsequence (LPS)Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
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Longest Palindromic SubstringGiven a string s, the task is to find the longest substring which is a palindrome. If there are multiple answers, then return the first appearing substring.Examples:Input: s = "forgeeksskeegfor" Output: "geeksskeeg"Explanation: There are several possible palindromic substrings like "kssk", "ss", "ee
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Palindrome
C Program to Check for Palindrome StringA string is said to be palindrome if the reverse of the string is the same as the string. In this article, we will learn how to check whether the given string is palindrome or not using C program.The simplest method to check for palindrome string is to reverse the given string and store it in a temp
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Check if a given string is a rotation of a palindromeGiven a string, check if it is a rotation of a palindrome. For example your function should return true for "aab" as it is a rotation of "aba". Examples: Input: str = "aaaad" Output: 1 // "aaaad" is a rotation of a palindrome "aadaa" Input: str = "abcd" Output: 0 // "abcd" is not a rotation of any p
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Check if characters of a given string can be rearranged to form a palindromeGiven a string, Check if the characters of the given string can be rearranged to form a palindrome. For example characters of "geeksogeeks" can be rearranged to form a palindrome "geeksoskeeg", but characters of "geeksforgeeks" cannot be rearranged to form a palindrome. Recommended PracticeAnagram P
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Online algorithm for checking palindrome in a streamGiven a stream of characters (characters are received one by one), write a function that prints 'Yes' if a character makes the complete string palindrome, else prints 'No'. Examples:Input: str[] = "abcba"Output: a Yes // "a" is palindrome b No // "ab" is not palindrome c No // "abc" is not palindrom
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Print all Palindromic Partitions of a String using Bit ManipulationGiven a string, find all possible palindromic partitions of a given string. Note that this problem is different from Palindrome Partitioning Problem, there the task was to find the partitioning with minimum cuts in input string. Here we need to print all possible partitions. Example: Input: nitinOut
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Minimum Characters to Add at Front for PalindromeGiven a string s, the task is to find the minimum number of characters to be added to the front of s to make it palindrome. A palindrome string is a sequence of characters that reads the same forward and backward. Examples: Input: s = "abc"Output: 2Explanation: We can make above string palindrome as
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Make largest palindrome by changing at most K-digitsYou are given a string s consisting of digits (0-9) and an integer k. Convert the string into a palindrome by changing at most k digits. If multiple palindromes are possible, return the lexicographically largest one. If it's impossible to form a palindrome with k changes, return "Not Possible".Examp
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Minimum Deletions to Make a String PalindromeGiven a string s of length n, the task is to remove or delete the minimum number of characters from the string so that the resultant string is a palindrome. Note: The order of characters should be maintained. Examples : Input : s = "aebcbda"Output : 2Explanation: Remove characters 'e' and 'd'. Resul
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Minimum insertions to form a palindrome with permutations allowedGiven a string of lowercase letters. Find minimum characters to be inserted in the string so that it can become palindrome. We can change the positions of characters in the string.Examples: Input: geeksforgeeksOutput: 2Explanation: geeksforgeeks can be changed as: geeksroforskeeg or geeksorfroskeeg
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