Given an integer **N** and a binary string consisting of **4*N** number of **0s** initially, the task is to flip the characters such that any two pair of indices of the string consisting of **1**s are neither co-prime nor the pair of indices can be divisible by each other. **Note:** Consider **1-based** indexing.

**Examples:**

Input:N = 3, S =“000000000000”Output:000000010101Explanation:In the modified string “000000010101”, the indices of 1s are {8, 10, 12}. In the above set of indices, there does not exist any pair of indices that are co-prime and divisible by each other.

Input:N = 2, S =“00000000”Output:00000101

**Approach:** The given problem can be solved based on the observation that if the characters are flipped at the positions **4*N, 4*N – 2, 4*N – 4, … **up to** N terms**, then there doesn’t exist any pair of indices that are divisible by each other and having GCD as **1**.

Below is the implementation of the above approach:

## C++

`#include <iostream>` `using` `namespace` `std;` `// Function to modify a string such` `// that there doesn't exist any pair` `// of indices consisting of 1s, whose` `// GCD is 1 and are divisible by each other` `void` `findString(` `char` `S[], ` `int` `N)` `{` ` ` `int` `strLen = 4 * N;` ` ` `// Flips characters at indices` ` ` `// 4N, 4N - 2, 4N - 4 .... upto N terms` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` `S[strLen - 1] = ` `'1'` `;` ` ` `strLen -= 2;` ` ` `}` ` ` `// Print the string` ` ` `for` `(` `int` `i = 0; i < 4 * N; i++) {` ` ` `cout << S[i];` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 2;` ` ` `char` `S[4 * N];` ` ` `// Initialize the string S` ` ` `for` `(` `int` `i = 0; i < 4 * N; i++)` ` ` `S[i] = ` `'0'` `;` ` ` `// function call` ` ` `findString(S, N);` ` ` `return` `0;` `}` `// This code is contributed by aditya7409.` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to modify a string such` ` ` `// that there doesn't exist any pair` ` ` `// of indices consisting of 1s, whose` ` ` `// GCD is 1 and are divisible by each other` ` ` `public` `static` `void` `findString(` `char` `S[], ` `int` `N)` ` ` `{` ` ` `int` `strLen = ` `4` `* N;` ` ` `// Flips characters at indices` ` ` `// 4N, 4N - 2, 4N - 4 .... upto N terms` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++) {` ` ` `S[strLen - ` `1` `] = ` `'1'` `;` ` ` `strLen -= ` `2` `;` ` ` `}` ` ` `// Print the string` ` ` `System.out.println(S);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `2` `;` ` ` `char` `S[] = ` `new` `char` `[` `4` `* N];` ` ` `// Initialize the string S` ` ` `for` `(` `int` `i = ` `0` `; i < ` `4` `* N; i++)` ` ` `S[i] = ` `'0'` `;` ` ` `findString(S, N);` ` ` `}` `}` |

## Python3

`# Python3 program for the above approach` `# Function to modify a string such` `# that there doesn't exist any pair` `# of indices consisting of 1s, whose` `# GCD is 1 and are divisible by each other` `def` `findString(S, N) :` ` ` `strLen ` `=` `4` `*` `N` ` ` `# Flips characters at indices` ` ` `# 4N, 4N - 2, 4N - 4 .... upto N terms` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `+` `1` `):` ` ` `S[strLen ` `-` `1` `] ` `=` `'1'` ` ` `strLen ` `-` `=` `2` ` ` ` ` `# Prthe string` ` ` `for` `i ` `in` `range` `(` `4` `*` `N):` ` ` `print` `(S[i], end ` `=` `"")` `# Driver code` `N ` `=` `2` `S ` `=` `[` `0` `] ` `*` `(` `4` `*` `N)` `# Initialize the string S` `for` `i ` `in` `range` `(` `4` `*` `N):` ` ` `S[i] ` `=` `'0'` ` ` `# function call` `findString(S, N)` `# This code is contributed by sanjoy_62.` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `public` `class` `GFG` `{` ` ` ` ` `// Function to modify a string such` ` ` `// that there doesn't exist any pair` ` ` `// of indices consisting of 1s, whose` ` ` `// GCD is 1 and are divisible by each other` ` ` `public` `static` `void` `findString(` `char` `[] S, ` `int` `N)` ` ` `{` ` ` `int` `strLen = 4 * N;` ` ` `// Flips characters at indices` ` ` `// 4N, 4N - 2, 4N - 4 .... upto N terms` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` `S[strLen - 1] = ` `'1'` `;` ` ` `strLen -= 2;` ` ` `}` ` ` `// Print the string` ` ` `Console.WriteLine(S);` ` ` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 2;` ` ` `char` `[] S = ` `new` `char` `[4 * N];` ` ` `// Initialize the string S` ` ` `for` `(` `int` `i = 0; i < 4 * N; i++)` ` ` `S[i] = ` `'0'` `;` ` ` `findString(S, N);` `}` `}` `// This code is contributed by souravghosh0416.` |

**Output:**

00000101

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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