Find the index at which bit must be set to maximise distance between next set bit

Given a binary array arr[]. The task is to find the position of any 0 in arr[] such that the distance between two set bits is maximized.

Examples

Input: arr = [1, 0, 0, 0, 1, 0, 1]
Output: 2
Explanation: Flip the bit at arr[2]

Input: arr = [1, 0, 0, 0]
Output: 3

Approach: The problem can be solved by finding the longest distance between adjacent set bits with some variation. Follow the steps below to solve the given problem.

• For all distances between adjacent set bits, find the maximum one and store its half as one of the required answers.
• Then find the distance between distance between 0 and the first set bit, and between index N-1 and last set bit.
• Find the overall maximum as the required answer.
• Print the answer found at the end.

Below is the implementation of the above approach.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum distance between any
// two set bits after flipping one bit
int maxDistToClosest1(vector<int>& arr)
{
    // The size of the array
    int n = arr.size(), ans = 0;

    int temp = 1, setbit = 0;

    // Iterate through the array
    for (int i = 1; i < n; i++) {
        if (arr[i] == 1) {

            if (setbit == 0 && arr[0] == 0)
                ans = max(ans, temp);
            else
                ans = max(ans, temp / 2);
            setbit = 1;
            temp = 0;
        }
        temp++;
    }
    ans = arr[n - 1] == 0 ? max(temp - 1, ans)
                          : max(temp / 2, ans);

    // Return the answer found
    return ans;
}

// Driver Code
int main()
{
    vector<int> arr = { 1, 0, 0, 0, 1, 0, 1 };

    // Function Call
    cout << maxDistToClosest1(arr);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find the maximum distance between any
  // two set bits after flipping one bit
  static int maxDistToClosest1(int arr[])
  {
    // The size of the array
    int n = arr.length, ans = 0;

    int temp = 1, setbit = 0;

    // Iterate through the array
    for (int i = 1; i < n; i++) {
      if (arr[i] == 1) {

        if (setbit == 0 && arr[0] == 0)
          ans = Math.max(ans, temp);
        else
          ans = Math.max(ans, temp / 2);
        setbit = 1;
        temp = 0;
      }
      temp++;
    }
    ans = arr[n - 1] == 0 ? Math.max(temp - 1, ans)
      : Math.max(temp / 2, ans);

    // Return the answer found
    return ans;
  }

  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 1, 0, 0, 0, 1, 0, 1 };

    // Function Call
    System.out.print(maxDistToClosest1(arr));

  }
}

// This code is contributed by hrithikgarg03188.

Python

# Python program for above approach

#  Function to find the maximum distance between any
#  two set bits after flipping one bit
def maxDistToClosest1(arr):

    # The size of the array
    n = len(arr)
    ans = 0

    temp = 1
    setbit = 0

    # Iterate through the array
    for i in range(1, n):
        if (arr[i] == 1):

            if (setbit == 0 and arr[0] == 0):
                ans = max(ans, temp)
            else:
                ans = max(ans, temp // 2)
            setbit = 1
            temp = 0

        temp +=1

    if(arr[n - 1] == 0):
        ans = max(temp - 1, ans)
    else:
        ans = max(temp // 2, ans)

    # Return the answer found
    return ans

# Driver Code
arr = [ 1, 0, 0, 0, 1, 0, 1 ]

# Function Call
print(maxDistToClosest1(arr))

# This code is contributed by Samim Hossain Mondal.

C#

// C# program for above approach
using System;
class GFG
{

  // Function to find the maximum distance between any
  // two set bits after flipping one bit
  static int maxDistToClosest1(int[] arr)
  {

    // The size of the array
    int n = arr.Length, ans = 0;

    int temp = 1, setbit = 0;

    // Iterate through the array
    for (int i = 1; i < n; i++) {
      if (arr[i] == 1) {

        if (setbit == 0 && arr[0] == 0)
          ans = Math.Max(ans, temp);
        else
          ans = Math.Max(ans, temp / 2);
        setbit = 1;
        temp = 0;
      }
      temp++;
    }
    ans = arr[n - 1] == 0 ? Math.Max(temp - 1, ans)
      : Math.Max(temp / 2, ans);

    // Return the answer found
    return ans;
  }

  // Driver Code
  public static int Main()
  {
    int[] arr = { 1, 0, 0, 0, 1, 0, 1 };

    // Function Call
    Console.Write(maxDistToClosest1(arr));
    return 0;
  }
}

// This code is contributed by Taranpreet

Javascript

<script>
    // JavaScript program for above approach

    // Function to find the maximum distance between any
    // two set bits after flipping one bit
    const maxDistToClosest1 = (arr) => {

        // The size of the array
        let n = arr.length, ans = 0;

        let temp = 1, setbit = 0;

        // Iterate through the array
        for (let i = 1; i < n; i++) {
            if (arr[i] == 1) {

                if (setbit == 0 && arr[0] == 0)
                    ans = Math.max(ans, temp);
                else
                    ans = Math.max(ans, parseInt(temp / 2));
                setbit = 1;
                temp = 0;
            }
            temp++;
        }
        ans = arr[n - 1] == 0 ? Math.max(temp - 1, ans)
            : Math.max(parseInt(temp / 2), ans);

        // Return the answer found
        return ans;
    }

    // Driver Code
    let arr = [1, 0, 0, 0, 1, 0, 1];

    // Function Call
    document.write(maxDistToClosest1(arr));

// This code is contributed by rakeshsahni

</script>

Output

2

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