Min Adjacent Swaps for Bracket Balancing
Last Updated :
02 Nov, 2024
You are given a string of 2N characters consisting of N '[' brackets and N ']' brackets. A string is considered balanced if it can be represented in the form S2[S1] where S1 and S2 are balanced strings. We can make an unbalanced string balanced by swapping adjacent characters. Calculate the minimum number of swaps necessary to make a string balanced.
Examples:
Input : []][][
Output : 2
First swap: Position 3 and 4
[][]][
Second swap: Position 5 and 6
[][][]
Input : [[][]]
Output : 0
The string is already balanced.
We can solve this problem by using greedy strategies. If the first X characters form a balanced string, we can neglect these characters and continue on. If we encounter a ']' before the required '[', then we must start swapping elements to balance the string.
Naive Approach
Initialize sum = 0 where sum stores result. Go through the string maintaining a count of the number of '[' brackets encountered. Reduce this count when we encounter a ']' character. If the count hits negative, then we must start balancing the string.
Let index 'i' represent the position we are at. We now move forward to the next '[' at index j. Increase sum by j - i. Move the '[' at position j, to position i, and shift all other characters to the right. Set the count back to 1 and continue traversing the string. In the end, 'sum' will have the required value.
Code-
C++
// C++ program to count swaps required to balance string
#include<bits/stdc++.h>
using namespace std;
// Function to calculate swaps required
int swapCount(string s)
{
//To store answer
int ans=0;
//To store count of '['
int count=0;
//Size of string
int n=s.size();
//Traverse over the string
for(int i=0;i<n;i++){
//When '[' encounters
if(s[i]=='['){count++;}
//when ']' encounters
else{count--;}
//When count becomes less than 0
if(count<0){
//Start searching for '[' from (i+1)th index
int j=i+1;
while(j<n){
//When jth index contains '['
if(s[j]=='['){break;}
j++;
}
//Increment answer
ans+=j-i;
//Set Count to 1 again
count=1;
//Bring character at jth position to ith position
//and shift all character from i to j-1
//towards right
char ch=s[j];
for(int k=j;k>i;k--){
s[k]=s[k-1];
}
s[i]=ch;
}
}
return ans;
}
// Driver code
int main()
{
string s = "[]][][";
cout << swapCount(s) << "\n";
s = "[[][]]";
cout << swapCount(s) << "\n";
return 0;
}
Java
// Java program to count swaps required to balance string
public class GFG {
// Function to calculate swaps required
static int swapCount(String s) {
//To store answer
int ans = 0;
//To store count of '['
int count = 0;
//Size of string
int n = s.length();
//Traverse over the string
for (int i = 0; i < n; i++) {
//When '[' encounters
if (s.charAt(i) == '[')
count++;
//when ']' encounters
else
count--;
//When count becomes less than 0
if (count < 0) {
//Start searching for '[' from (i+1)th index
int j = i + 1;
while (j < n) {
//When jth index contains '['
if (s.charAt(j) == '[')
break;
j++;
}
//Increment answer
ans += j - i;
//Set Count to 1 again
count = 1;
//Bring character at jth position to ith position
//and shift all character from i to j-1
//towards right
char ch = s.charAt(j);
StringBuilder newString = new StringBuilder(s);
for (int k = j; k > i; k--) {
newString.setCharAt(k, s.charAt(k - 1));
}
newString.setCharAt(i, ch);
s = newString.toString();
}
}
return ans;
}
// Driver code
public static void main(String[] args) {
String s = "[]][][";
System.out.println(swapCount(s));
s = "[[][]]";
System.out.println(swapCount(s));
}
}
Python
def swap_count(s):
# To store the answer
ans = 0
# To store the count of '['
count = 0
# Size of the string
n = len(s)
# Traverse over the string
for i in range(n):
# When '[' encounters
if s[i] == '[':
count += 1
# When ']' encounters
else:
count -= 1
# When count becomes less than 0
if count < 0:
# Start searching for '[' from (i+1)th index
j = i + 1
while j < n:
# When jth index contains '['
if s[j] == '[':
break
j += 1
# Increment the answer
ans += j - i
# Set count to 1 again
count = 1
# Bring the character at jth position to ith position
# and shift all characters from i to j-1
# towards the right
ch = s[j]
for k in range(j, i, -1):
s[k] = s[k - 1]
s[i] = ch
return ans
# Driver code
if __name__ == "__main__":
s = "[]][]["
print(swap_count(list(s)))
s = "[[][]]"
print(swap_count(list(s)))
C#
using System;
class Program
{
// Function to calculate swaps required
static int SwapCount(string s)
{
// To store answer
int ans = 0;
// To store count of '['
int count = 0;
// Size of string
int n = s.Length;
// Traverse over the string
for (int i = 0; i < n; i++)
{
// When '[' encounters
if (s[i] == '[')
{
count++;
}
// When ']' encounters
else
{
count--;
}
// When count becomes less than 0
if (count < 0)
{
// Start searching for '[' from (i+1)th index
int j = i + 1;
while (j < n)
{
// When jth index contains '['
if (s[j] == '[')
{
break;
}
j++;
}
// Increment answer
ans += j - i;
// Set Count to 1 again
count = 1;
// Bring character at jth position to ith position
// and shift all characters from i to j-1
// towards right
char ch = s[j];
for (int k = j; k > i; k--)
{
s = s.Remove(k, 1);
s = s.Insert(k, s[k - 1].ToString());
}
s = s.Remove(i, 1);
s = s.Insert(i, ch.ToString());
}
}
return ans;
}
// Driver code
static void Main(string[] args)
{
string s = "[]][][";
Console.WriteLine(SwapCount(s));
s = "[[][]]";
Console.WriteLine(SwapCount(s));
}
}
JavaScript
function GFG(s) {
// To store answer
let ans = 0;
// To store count of '['
let count = 0;
// Traverse over the string
for (let i = 0; i < s.length; i++) {
// When '[' encounters
if (s[i] === '[') {
count++;
}
// When ']' encounters
else {
count--;
}
// When count becomes less than 0
if (count < 0) {
// Start searching for
// '[' from the beginning
for (let j = 0; j < i; j++) {
// When jth index contains '['
if (s[j] === '[') {
ans += i - j;
// Swap characters to balance the string
let temp = s.substring(0, j) + s[i] + s.substring(j + 1, i) + s[j] + s.substring(i + 1);
s = temp;
break;
}
}
// Reset the count
count = 1;
}
}
return ans;
}
// Driver code
let s1 = "[]][][";
console.log(GFG(s1)); // Output: 2
let s2 = "[[][]]";
console.log(GFG(s2)); // Output: 0
Time Complexity = O(N^2), one loop is for traversing the string and another loop in finding the next '[' when the count becomes less than 0 and making the string ready for the next step
Extra Space = O(1), because no extra space has been used
Optimized approach
We can initially go through the string and store the positions of '[' in a vector say 'pos'. Initialize 'p' to 0. We shall use p to traverse the vector 'pos'. Similar to the naive approach, we maintain a count of encountered '[' brackets. When we encounter a '[' we increase the count and increase 'p' by 1. When we encounter a ']' we decrease the count. If the count ever goes negative, this means we must start swapping. The element pos[p] tells us the index of the next '['. We increase the sum by pos[p] - i, where i is the current index. We can swap the elements in the current index and pos[p] and reset the count to 1 and increment p so that it pos[p] indicates to the next '['.
Since we have converted a step that was O(N) in the naive approach, to an O(1) step, our new time complexity reduces.
Time Complexity = O(N)
Extra Space = O(N)
C++
// C++ program to count swaps required to balance string
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Function to calculate swaps required
long swapCount(string s)
{
// Keep track of '['
vector<int> pos;
for (int i = 0; i < s.length(); ++i)
if (s[i] == '[')
pos.push_back(i);
int count = 0; // To count number of encountered '['
int p = 0; // To track position of next '[' in pos
long sum = 0; // To store result
for (int i = 0; i < s.length(); ++i)
{
// Increment count and move p to next position
if (s[i] == '[')
{
++count;
++p;
}
else if (s[i] == ']')
--count;
// We have encountered an unbalanced part of string
if (count < 0)
{
// Increment sum by number of swaps required
// i.e. position of next '[' - current position
sum += pos[p] - i;
swap(s[i], s[pos[p]]);
++p;
// Reset count to 1
count = 1;
}
}
return sum;
}
// Driver code
int main()
{
string s = "[]][][";
cout << swapCount(s) << "\n";
s = "[[][]]";
cout << swapCount(s) << "\n";
return 0;
}
Java
// Java program to count swaps
// required to balance string
import java.util.*;
class GFG{
// Function to calculate swaps required
public static long swapCount(String s)
{
// Keep track of '['
Vector<Integer> pos = new Vector<Integer>();
for(int i = 0; i < s.length(); ++i)
if (s.charAt(i) == '[')
pos.add(i);
// To count number of encountered '['
int count = 0;
// To track position of next '[' in pos
int p = 0;
// To store result
long sum = 0;
char[] S = s.toCharArray();
for(int i = 0; i < s.length(); ++i)
{
// Increment count and move p
// to next position
if (S[i] == '[')
{
++count;
++p;
}
else if (S[i] == ']')
--count;
// We have encountered an
// unbalanced part of string
if (count < 0)
{
// Increment sum by number of
// swaps required i.e. position
// of next '[' - current position
sum += pos.get(p) - i;
char temp = S[i];
S[i] = S[pos.get(p)];
S[pos.get(p)] = temp;
++p;
// Reset count to 1
count = 1;
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
String s = "[]][][";
System.out.println(swapCount(s));
s = "[[][]]";
System.out.println(swapCount(s));
}
}
// This code is contributed by divyesh072019
Python
# Python3 Program to count
# swaps required to balance
# string
# Function to calculate
# swaps required
def swapCount(s):
# Keep track of '['
pos = []
for i in range(len(s)):
if(s[i] == '['):
pos.append(i)
# To count number
# of encountered '['
count = 0
# To track position
# of next '[' in pos
p = 0
# To store result
sum = 0
s = list(s)
for i in range(len(s)):
# Increment count and
# move p to next position
if(s[i] == '['):
count += 1
p += 1
elif(s[i] == ']'):
count -= 1
# We have encountered an
# unbalanced part of string
if(count < 0):
# Increment sum by number
# of swaps required
# i.e. position of next
# '[' - current position
sum += pos[p] - i
s[i], s[pos[p]] = (s[pos[p]],
s[i])
p += 1
# Reset count to 1
count = 1
return sum
# Driver code
s = "[]][]["
print(swapCount(s))
s = "[[][]]"
print(swapCount(s))
# This code is contributed by avanitrachhadiya2155
C#
// C# program to count swaps
// required to balance string
using System.IO;
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function to calculate swaps required
static long swapCount(string s)
{
// Keep track of '['
List<int> pos = new List<int>();
for(int i = 0; i < s.Length; i++)
{
if (s[i] == '[')
{
pos.Add(i);
}
}
// To count number of encountered '['
int count = 0;
// To track position of next '[' in pos
int p = 0;
// To store result
long sum = 0;
char[] S = s.ToCharArray();
for(int i = 0; i < S.Length; i++)
{
// Increment count and move p
// to next position
if (S[i] == '[')
{
++count;
++p;
}
else if (S[i] == ']')
{
--count;
}
// We have encountered an
// unbalanced part of string
if (count < 0)
{
// Increment sum by number of
// swaps required i.e. position
// of next '[' - current position
sum += pos[p]-i;
char temp = S[i];
S[i] = S[pos[p]];
S[pos[p]] = temp;
++p;
// Reset count to 1
count = 1;
}
}
return sum;
}
// Driver code
static void Main()
{
string s = "[]][][";
Console.WriteLine(swapCount(s));
s = "[[][]]";
Console.WriteLine(swapCount(s));
}
}
// This code is contributed by rag2127
JavaScript
// JavaScript program to count swaps
// required to balance string
// Function to calculate swaps required
function swapCount(s)
{
// Keep track of '['
let pos = [];
for(let i = 0; i < s.length; ++i)
if (s[i] == '[')
pos.push(i);
// To count number of encountered '['
let count = 0;
// To track position of next '[' in pos
let p = 0;
// To store result
let sum = 0;
let S = s.split('');
for(let i = 0; i < s.length; ++i)
{
// Increment count and move p
// to next position
if (S[i] == '[')
{
++count;
++p;
}
else if (S[i] == ']')
--count;
// We have encountered an
// unbalanced part of string
if (count < 0)
{
// Increment sum by number of
// swaps required i.e. position
// of next '[' - current position
sum += pos[p] - i;
let temp = S[i];
S[i] = S[pos[p]];
S[pos[p]] = temp;
++p;
// Reset count to 1
count = 1;
}
}
return sum;
}
// Driver Code
let s = "[]][][";
console.log(swapCount(s) + "<br/>");
s = "[[][]]";
console.log(swapCount(s));
Time Complexity: O(N)
Auxiliary Space: O(N)
Another Method:
We can do without having to store the positions of '['.
Below is the implementation :
C++
// C++ program to count swaps required
// to balance string
#include <bits/stdc++.h>
using namespace std;
long swapCount(string chars)
{
// Stores total number of Left and
// Right brackets encountered
int countLeft = 0, countRight = 0;
// swap stores the number of swaps
// required imbalance maintains
// the number of imbalance pair
int swap = 0 , imbalance = 0;
for(int i = 0; i < chars.length(); i++)
{
if (chars[i] == '[')
{
// Increment count of Left bracket
countLeft++;
if (imbalance > 0)
{
// swaps count is last swap count + total
// number imbalanced brackets
swap += imbalance;
// imbalance decremented by 1 as it solved
// only one imbalance of Left and Right
imbalance--;
}
}
else if(chars[i] == ']' )
{
// Increment count of Right bracket
countRight++;
// imbalance is reset to current difference
// between Left and Right brackets
imbalance = (countRight - countLeft);
}
}
return swap;
}
// Driver code
int main()
{
string s = "[]][][";
cout << swapCount(s) << endl;
s = "[[][]]";
cout << swapCount(s) << endl;
return 0;
}
// This code is contributed by divyeshrabadiya07
Java
// Java Program to count swaps required to balance string
public class BalanceParan
{
static long swapCount(String s)
{
char[] chars = s.toCharArray();
// stores total number of Left and Right
// brackets encountered
int countLeft = 0, countRight = 0;
// swap stores the number of swaps required
//imbalance maintains the number of imbalance pair
int swap = 0 , imbalance = 0;
for(int i =0; i< chars.length; i++)
{
if(chars[i] == '[')
{
// increment count of Left bracket
countLeft++;
if(imbalance > 0)
{
// swaps count is last swap count + total
// number imbalanced brackets
swap += imbalance;
// imbalance decremented by 1 as it solved
// only one imbalance of Left and Right
imbalance--;
}
} else if(chars[i] == ']' )
{
// increment count of Right bracket
countRight++;
// imbalance is reset to current difference
// between Left and Right brackets
imbalance = (countRight-countLeft);
}
}
return swap;
}
// Driver code
public static void main(String args[])
{
String s = "[]][][";
System.out.println(swapCount(s) );
s = "[[][]]";
System.out.println(swapCount(s) );
}
}
// This code is contributed by Janmejaya Das.
Python
# Python3 program to count swaps required to
# balance string
def swapCount(s):
# Swap stores the number of swaps
# required imbalance maintains the
# number of imbalance pair
swap = 0
imbalance = 0;
for i in s:
if i == '[':
# Decrement the imbalance
imbalance -= 1
else:
# Increment imbalance
imbalance += 1
if imbalance > 0:
swap += imbalance
return swap
# Driver code
s = "[]][][";
print(swapCount(s))
s = "[[][]]";
print(swapCount(s))
# This code is contributed by Prateek Gupta and improved by Anvesh Govind Saxena
C#
// C# Program to count swaps required
// to balance string
using System;
class GFG
{
public static long swapCount(string s)
{
char[] chars = s.ToCharArray();
// stores the total number of Left and
// Right brackets encountered
int countLeft = 0, countRight = 0;
// swap stores the number of swaps
// required imbalance maintains the
// number of imbalance pair
int swap = 0, imbalance = 0;
for (int i = 0; i < chars.Length; i++)
{
if (chars[i] == '[')
{
// increment count of Left bracket
countLeft++;
if (imbalance > 0)
{
// swaps count is last swap count + total
// number imbalanced brackets
swap += imbalance;
// imbalance decremented by 1 as it solved
// only one imbalance of Left and Right
imbalance--;
}
}
else if (chars[i] == ']')
{
// increment count of Right bracket
countRight++;
// imbalance is reset to current difference
// between Left and Right brackets
imbalance = (countRight - countLeft);
}
}
return swap;
}
// Driver code
public static void Main(string[] args)
{
string s = "[]][][";
Console.WriteLine(swapCount(s));
s = "[[][]]";
Console.WriteLine(swapCount(s));
}
}
// This code is contributed by Shrikant13
JavaScript
// Javascript Program to count swaps required
// to balance string
function swapCount(s)
{
let chars = s.split('');
// stores the total number of Left and
// Right brackets encountered
let countLeft = 0, countRight = 0;
// swap stores the number of swaps
// required imbalance maintains the
// number of imbalance pair
let swap = 0, imbalance = 0;
for (let i = 0; i < chars.length; i++)
{
if (chars[i] == '[')
{
// increment count of Left bracket
countLeft++;
if (imbalance > 0)
{
// swaps count is last swap count + total
// number imbalanced brackets
swap += imbalance;
// imbalance decremented by 1 as it solved
// only one imbalance of Left and Right
imbalance--;
}
}
else if (chars[i] == ']')
{
// increment count of Right bracket
countRight++;
// imbalance is reset to current difference
// between Left and Right brackets
imbalance = (countRight - countLeft);
}
}
return swap;
}
let s = "[]][][";
console.log(swapCount(s) + "</br>");
s = "[[][]]";
console.log(swapCount(s));
// This code is contributed by suresh07.
Time Complexity :O(N)
Auxiliary Space : O(1)
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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 (â\0â). 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 = â001â 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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K'th bit in a binary representation with n iterationsGiven a decimal number m. Consider its binary representation string and apply n iterations. In each iteration, replace the character 0 with the string 01, and 1 with 10. Find the kth (1-based indexing) character in the string after the nth iterationExamples: Input: m = 5, n = 2, k = 5Output: 0Explan
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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 two subsequences.[abba], [abba]Input: txt = banana, pat = banOutput: 3Explanation
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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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