K'th bit in a binary representation with n iterations
Last Updated :
20 Jun, 2025
Given 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 iteration
Examples:
Input: m = 5, n = 2, k = 5
Output: 0
Explanation: Binary representation of m is "101", after one iteration binary representation will be "100110", and after second iteration binary representation will be "100101101001".
Input: m = 5, n = 2, k = 1
Output: 1
Explanation: Binary representation of m is "101", after one iteration binary representation will be "100110", and after second iteration binary representation will be "100101101001".
[Naive Approach] Stimulating Each Iteration - O(2^n) Time and O(n) Space
The idea is to do as told in the problem, at every iteration we update the binary representation(i.e. update 0 to 01 and 1 to 10).
Step-By-Step Implementation:
- Change the given number into a binary and store it in string s.
- Run loop n times. Run another loop for string length s to convert 0 to “01” and 1 to “10” and store in a temporary string s1. After completion of each iteration, assign string s1 to s.
- Finally, return the value of the kth index in string s.
C++
// C++ Program to find kth character in
// a binary string.
#include <bits/stdc++.h>
using namespace std;
// Function to store binary Representation
string binaryRep(int m)
{
string s = "";
while (m)
{
int tmp = m % 2;
s += tmp + '0';
m = m / 2;
}
reverse(s.begin(), s.end());
return s;
}
// Function to find kth character
int findKthChar(int n, int m, int k)
{
string s = binaryRep(m);
string s1 = "";
for (int x = 0; x < n; x++)
{
for (int y = 0; y < s.length(); y++)
{
if (s[y] == '1')
s1 += "10";
else
s1 += "01";
}
// Assign s1 string in s string
s = s1;
s1 = "";
}
return s[k] - '0';
}
// Driver Function
int main()
{
int m = 5, n = 2, k = 8;
cout << findKthChar(n, m, k);
return 0;
}
// Java Program to find kth character in a binary string.
class Main {
// Function to store binary Representation
static String binaryRep(int m) {
StringBuilder s = new StringBuilder();
while (m > 0) {
int tmp = m % 2;
s.append(tmp);
m = m / 2;
}
return s.reverse().toString();
}
// Function to find kth character
static int findKthChar(int n, int m, int k) {
String s = binaryRep(m);
StringBuilder s1 = new StringBuilder();
for (int x = 0; x < n; x++) {
for (int y = 0; y < s.length(); y++) {
if (s.charAt(y) == '1')
s1.append("10");
else
s1.append("01");
}
// Assign s1 string in s string
s = s1.toString();
s1.setLength(0);
}
return s.charAt(k) - '0';
}
// Driver Function
public static void main(String[] args) {
int m = 5, n = 2, k = 8;
System.out.println(findKthChar(n, m, k));
}
}
# Python Program to find kth character in a binary string.
# Function to store binary Representation
def binaryRep(m):
s = ""
while m:
tmp = m % 2
s += str(tmp)
m //= 2
return s[::-1]
# Function to find kth character
def findKthChar(n, m, k):
s = binaryRep(m)
for x in range(n):
s1 = ""
for y in range(len(s)):
if s[y] == '1':
s1 += "10"
else:
s1 += "01"
# Assign s1 string in s string
s = s1
return int(s[k])
# Driver Function
m = 5
n = 2
k = 8
print(findKthChar(n, m, k))
// C# Program to find kth character in a binary string.
using System;
class Program {
// Function to store binary Representation
static string BinaryRep(int m) {
string s = "";
while (m > 0) {
int tmp = m % 2;
s += tmp.ToString();
m /= 2;
}
char[] charArray = s.ToCharArray();
Array.Reverse(charArray);
return new string(charArray);
}
// Function to find kth character
static int FindKthChar(int n, int m, int k) {
string s = BinaryRep(m);
for (int x = 0; x < n; x++) {
string s1 = "";
for (int y = 0; y < s.Length; y++) {
if (s[y] == '1')
s1 += "10";
else
s1 += "01";
}
// Assign s1 string in s string
s = s1;
}
return s[k] - '0';
}
// Driver Function
static void Main() {
int m = 5, n = 2, k = 8;
Console.WriteLine(FindKthChar(n, m, k));
}
}
// JavaScript Program to find kth character in a binary string.
// Function to store binary Representation
function binaryRep(m) {
let s = "";
while (m > 0) {
let tmp = m % 2;
s += tmp;
m = Math.floor(m / 2);
}
return s.split('').reverse().join('');
}
// Function to find kth character
function findKthChar(n, m, k) {
let s = binaryRep(m);
for (let x = 0; x < n; x++) {
let s1 = "";
for (let y = 0; y < s.length; y++) {
if (s[y] === '1')
s1 += "10";
else
s1 += "01";
}
// Assign s1 string in s string
s = s1;
}
return parseInt(s[k]);
}
// Driver Function
let m = 5, n = 2, k = 8;
console.log(findKthChar(n, m, k));
[Expected Approach 1] Dividing String into Blocks - O(log(n)) Time and O(1) Space
The idea is to avoid generating the full binary string after each iteration, since its size doubles every time. Instead, we identify the original bit in m that gives rise to the k-th character after n iterations. The key observation is that the number of bit flips in a path from root to k-th character is equal to the number of set bits in offset, and parity of this count decides the final bit.
The first step will be to find which block the k-th character will be after n iterations are performed. In the n'th iteration distance between any two consecutive characters initially will always be equal to 2^n.
Consider input number m = 5
0th iteration: 101, distance = 0
1st iteration: 10 01 10, distance = 2
2nd iteration: 1001 0110 1001, distance = 4
3rd iteration: 10010110 01101001 10010110, distance = 8
We consider every bit of the input number as the start of a block and aim to determine the block number where the k-th bit will be located after n iterations. T
Finding the Block & Offset Within Block
- With each iteration, the distance between bits doubles.
- After
n iterations, the bits are spaced by 2ⁿ. - So, to find which block contains the k-th bit, we compute: {\scriptsize \text{Block number} = \left\lfloor \frac{k}{2^n} \right\rfloor}
- Example:
If k = 5 and n = 3, then {\scriptsize
\text{Block number} = \left\lfloor \frac{5}{8} \right\rfloor = 0
}
- The position of the k-th bit within the selected block is given by the remainder: {\scriptsize
\text{Remaining} = k \bmod 2^n
}
=
k % 2n
Finding k'th Bit
This offset can be thought of as a path in a binary tree formed by the repeated expansions. Each bit in the binary representation of this offset represents a decision point, a 1 indicates a flip in the bit. So, we count the number of 1s in the offset; this count gives us the number of bit flips applied to the base bit to reach the k-th position.
Finally, we use the parity of this flip count:
- If the base bit
b is 0, we start with '0'. An even number of flips keeps it '0', while an odd number of flips changes it to '1'. - If the base bit
b is 1, we start with '1'. An even number of flips keeps it '1', while an odd number of flips changes it to '0'.
This flipping behavior is directly derived from the rules of expansion (0 → 01, 1 → 10), where each level toggles the bit depending on the path taken. Using this approach, we determine the final k-th bit efficiently in logarithmic time without constructing the full expanded string.
C++
// C++ program to find k-th character after
// n expansions
#include <bits/stdc++.h>
using namespace std;
// Function to count number of set bits (Hamming weight)
int countBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to get k-th character after n expansions
char KthCharacter(int m, int n, int k) {
int len = 1 << n;
int bit_index = (k - 1) / len;
// Get total bits in m (MSB to LSB)
int total_bits = 31 - __builtin_clz(m);
int shift = total_bits - bit_index;
int b = (m >> shift) & 1;
int offset = (k - 1) % len;
// Count set bits (parity) in offset
int ones = countBits(offset);
// Flip if parity is odd
if (b == 0) {
return (ones % 2 == 0) ? '0' : '1';
} else {
return (ones % 2 == 0) ? '1' : '0';
}
}
// Driver code
int main() {
int m = 5, k = 5, n = 3;
cout << KthCharacter(m, n, k) << endl;
return 0;
}
// Java program to find k-th character after
// n expansions
class GfG {
// Function to count number of set bits (Hamming weight)
static int countBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to get k-th character after n expansions
static char KthCharacter(int m, int n, int k) {
int len = 1 << n;
int bit_index = (k - 1) / len;
// Get total bits in m (MSB to LSB)
int total_bits = 31 - Integer.numberOfLeadingZeros(m);
int shift = total_bits - bit_index;
int b = (m >> shift) & 1;
int offset = (k - 1) % len;
// Count set bits (parity) in offset
int ones = countBits(offset);
// Flip if parity is odd
if (b == 0) {
return (ones % 2 == 0) ? '0' : '1';
} else {
return (ones % 2 == 0) ? '1' : '0';
}
}
// Driver code
public static void main(String[] args) {
int m = 5, k = 5, n = 3;
System.out.println(KthCharacter(m, n, k));
}
}
# Python program to find k-th character after
# n expansions
# Function to count number of set bits (Hamming weight)
def countBits(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
# Function to get k-th character after n expansions
def KthCharacter(m, n, k):
len_ = 1 << n
bit_index = (k - 1) // len_
# Get total bits in m (MSB to LSB)
total_bits = m.bit_length() - 1
shift = total_bits - bit_index
b = (m >> shift) & 1
offset = (k - 1) % len_
# Count set bits (parity) in offset
ones = countBits(offset)
# Flip if parity is odd
if b == 0:
return '0' if ones % 2 == 0 else '1'
else:
return '1' if ones % 2 == 0 else '0'
# Driver code
if __name__ == "__main__":
m = 5
k = 5
n = 3
print(KthCharacter(m, n, k))
// C# program to find k-th character after
// n expansions
using System;
class GfG {
// Function to count number of set bits (Hamming weight)
static int countBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to get total bits (like bit_length)
static int totalBits(int m) {
int bits = 0;
while (m > 0) {
m >>= 1;
bits++;
}
return bits;
}
// Function to get k-th character after n expansions
static char KthCharacter(int m, int n, int k) {
int len = 1 << n;
int bit_index = (k - 1) / len;
// Get total bits in m (MSB to LSB)
int total_bits = totalBits(m) - 1;
int shift = total_bits - bit_index;
int b = (m >> shift) & 1;
int offset = (k - 1) % len;
// Count set bits (parity) in offset
int ones = countBits(offset);
// Flip if parity is odd
if (b == 0) {
return (ones % 2 == 0) ? '0' : '1';
} else {
return (ones % 2 == 0) ? '1' : '0';
}
}
// Driver code
public static void Main() {
int m = 5, k = 5, n = 3;
Console.WriteLine(KthCharacter(m, n, k));
}
}
// JavaScript program to find k-th character after
// n expansions
// Function to count number of set bits (Hamming weight)
function countBits(x) {
let cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to get k-th character after n expansions
function KthCharacter(m, n, k) {
let len = 1 << n;
let bit_index = Math.floor((k - 1) / len);
// Get total bits in m (MSB to LSB)
let total_bits = 31 - Math.clz32(m);
let shift = total_bits - bit_index;
let b = (m >> shift) & 1;
let offset = (k - 1) % len;
// Count set bits (parity) in offset
let ones = countBits(offset);
// Flip if parity is odd
if (b === 0) {
return (ones % 2 === 0) ? '0' : '1';
} else {
return (ones % 2 === 0) ? '1' : '0';
}
}
// Driver code
let m = 5, k = 5, n = 3;
console.log(KthCharacter(m, n, k));
[Expected Approach 2] Using Bitset - O(1) Time and O(1) Space
The idea is to avoid generating the full binary string after each iteration, since its size doubles every time. Instead, we identify the original bit in m that gives rise to the k-th character after n iterations. The key observation is that the number of bit flips in a path from root to k-th character is equal to the number of set bits in offset, and parity of this count decides the final bit.
C++
// C++ Code to find kth chracter after
// n iterations
#include <bits/stdc++.h>
using namespace std;
// Function to count number of set
// bits (used to compute parity)
int countSetBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to find the k-th character
// after n iterations
char KthCharacter(int m, int n, int k) {
bitset<32> binary(m);
int totalBits = 0;
// Count number of bits in m (from MSB)
for (int i = 31; i >= 0; i--) {
if (binary[i] == 1) {
totalBits = i + 1;
break;
}
}
int blockLength = 1 << n;
int blockIndex = (k - 1) / blockLength;
int offset = (k - 1) % blockLength;
// If m has fewer bits than blockIndex, bit is 0
int root = (blockIndex < totalBits) ? binary[totalBits - blockIndex - 1] : 0;
int flips = countSetBits(offset);
// Flip if parity is odd
if (flips % 2 == 0) {
return root ? '1' : '0';
} else {
return root ? '0' : '1';
}
}
// Driver code
int main() {
int m = 5, n = 2, k = 5;
cout << KthCharacter(m, n, k) << endl;
return 0;
}
// Java Code to find kth character after
// n iterations
class GfG {
// Function to count number of set
// bits (used to compute parity)
static int countSetBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to find the k-th character
// after n iterations
static char KthCharacter(int m, int n, int k) {
int[] binary = new int[32];
for (int i = 0; i < 32; i++) {
binary[i] = (m >> i) & 1;
}
int totalBits = 0;
// Count number of bits in m (from MSB)
for (int i = 31; i >= 0; i--) {
if (binary[i] == 1) {
totalBits = i + 1;
break;
}
}
int blockLength = 1 << n;
int blockIndex = (k - 1) / blockLength;
int offset = (k - 1) % blockLength;
// If m has fewer bits than blockIndex, bit is 0
int root = (blockIndex < totalBits) ?
binary[totalBits - blockIndex - 1] : 0;
int flips = countSetBits(offset);
// Flip if parity is odd
if (flips % 2 == 0) {
return root == 1 ? '1' : '0';
} else {
return root == 1 ? '0' : '1';
}
}
public static void main(String[] args) {
int m = 5, n = 2, k = 5;
System.out.println(KthCharacter(m, n, k));
}
}
# Python Code to find kth character after
# n iterations
# Function to count number of set
# bits (used to compute parity)
def countSetBits(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
# Function to find the k-th character
# after n iterations
def KthCharacter(m, n, k):
binary = [0] * 32
for i in range(32):
binary[i] = (m >> i) & 1
totalBits = 0
# Count number of bits in m (from MSB)
for i in range(31, -1, -1):
if binary[i] == 1:
totalBits = i + 1
break
blockLength = 1 << n
blockIndex = (k - 1) // blockLength
offset = (k - 1) % blockLength
# If m has fewer bits than blockIndex, bit is 0
root = binary[totalBits - blockIndex - 1] if blockIndex < totalBits else 0
flips = countSetBits(offset)
# Flip if parity is odd
if flips % 2 == 0:
return '1' if root else '0'
else:
return '0' if root else '1'
if __name__ == "__main__":
m, n, k = 5, 2, 5
print(KthCharacter(m, n, k))
// C# Code to find kth character after
// n iterations
using System;
class GfG {
// Function to count number of set
// bits (used to compute parity)
static int countSetBits(int x) {
int cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to find the k-th character
// after n iterations
static char KthCharacter(int m, int n, int k) {
int[] binary = new int[32];
for (int i = 0; i < 32; i++) {
binary[i] = (m >> i) & 1;
}
int totalBits = 0;
// Count number of bits in m (from MSB)
for (int i = 31; i >= 0; i--) {
if (binary[i] == 1) {
totalBits = i + 1;
break;
}
}
int blockLength = 1 << n;
int blockIndex = (k - 1) / blockLength;
int offset = (k - 1) % blockLength;
// If m has fewer bits than blockIndex, bit is 0
int root = (blockIndex < totalBits) ?
binary[totalBits - blockIndex - 1] : 0;
int flips = countSetBits(offset);
// Flip if parity is odd
if (flips % 2 == 0) {
return root == 1 ? '1' : '0';
} else {
return root == 1 ? '0' : '1';
}
}
static void Main() {
int m = 5, n = 2, k = 5;
Console.WriteLine(KthCharacter(m, n, k));
}
}
// JavaScript Code to find kth character after
// n iterations
// Function to count number of set
// bits (used to compute parity)
function countSetBits(x) {
let cnt = 0;
while (x > 0) {
cnt += x & 1;
x >>= 1;
}
return cnt;
}
// Function to find the k-th character
// after n iterations
function KthCharacter(m, n, k) {
let binary = new Array(32).fill(0);
for (let i = 0; i < 32; i++) {
binary[i] = (m >> i) & 1;
}
let totalBits = 0;
// Count number of bits in m (from MSB)
for (let i = 31; i >= 0; i--) {
if (binary[i] === 1) {
totalBits = i + 1;
break;
}
}
let blockLength = 1 << n;
let blockIndex = Math.floor((k - 1) / blockLength);
let offset = (k - 1) % blockLength;
// If m has fewer bits than blockIndex, bit is 0
let root = (blockIndex < totalBits) ?
binary[totalBits - blockIndex - 1] : 0;
let flips = countSetBits(offset);
// Flip if parity is odd
if (flips % 2 === 0) {
return root ? '1' : '0';
} else {
return root ? '0' : '1';
}
}
// Driver code
let m = 5, n = 2, k = 5;
console.log(KthCharacter(m, n, k));
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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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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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