找到必须设置位以最大化下一个设置位之间的距离的索引

给定一个二进制数组arr[] 。任务是在arr[]中找到任何0的位置，以使两个设置位之间的距离最大化。

例子

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

方法：这个问题可以通过找到一些变化的相邻设置位之间的最长距离来解决。请按照以下步骤解决给定的问题。

• 对于相邻设置位之间的所有距离，找到最大的一个并将其一半存储为所需答案之一。
• 然后找到0和第一个设置位之间的距离，以及索引N-1和最后一个设置位之间的距离。
• 找到总体最大值作为所需答案。
• 打印最后找到的答案。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find the maximum distance between any
// two set bits after flipping one bit
int maxDistToClosest1(vector& 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 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

# Pyhton 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

    // 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
 

输出

2

时间复杂度： O(N)
辅助空间： O(1)