翻转M 0后，最大化任意两个连续1之间的距离

给定一个二进制数组的大小，该数组仅由0和n组成，该整数由0和n组成；整数m是从0到1允许的翻转次数；任务是将m 0翻转为1后，使任意两个连续的1之间的距离最大。

例子：

Input: n = 5, m = 3
Output: 1
Explanation:
The initial array is arr = { 0, 0, 0, 0, 0 },
The final array is arr = { 1, 0, 1, 0, 1 },
So distance between two consecutive 1’s is 2.

Input: n = 9, m = 3
Output: 4
Explanation:
The initial array is arr = { 0, 0, 0, 0, 0, 0, 0, 0, 0 },
The final array is arr = { 1, 0, 0, 0, 1, 0, 0, 0, 1 },
so distance between two consecutive 1’s 4.

方法：

• 我们可以简单地对任意两个连续的数字之间的距离进行二进制搜索，并检查是否可以将m个零转换为1。
• 首先，我们设置low = 1，high = n – 1
• 然后检查中=(低+高)/ 2是否合适。
• 如果更新后的答案是中，则降低上限=中– 1。

下面是上述方法的实现：

CPP

// C++ program to Maximize distance between
// any two consecutive 1's after flipping M 0's
  
#include 
using namespace std;
  
// Function to return the count
bool check(int arr[], int n, int m, int d)
{
    // Flipping zeros at distance "d"
    int i = 0;
    while (i < n && m > 0) {
        m--;
        i += d;
    }
  
    return m == 0 ? true : false;
}
  
// Function to implement
// binary search
int maximumDistance(int arr[], int n, int m)
{
  
    int low = 1, high = n - 1;
    int ans = 0;
  
    while (low <= high) {
  
        int mid = (low + high) / 2;
  
        // Check for valid distance i.e mid
        bool flag = check(arr, n, m, mid);
  
        if (flag) {
            ans = mid;
            low = mid + 1;
        }
        else {
            high = mid - 1;
        }
    }
  
    return ans;
}
  
// Driver code
int main()
{
  
    int n = 5, m = 3;
    int arr[n] = { 0 };
  
    cout << maximumDistance(arr, n, m);
  
    return 0;
}

Java

// Java program to Maximize distance between
// any two consecutive 1's after flipping M 0's
  
class GFG 
{
  
    // Function to return the count
    static boolean check(int arr[], int n, int m, int d)
    {
          
        // Flipping zeros at distance "d"
        int i = 0;
        while (i < n && m > 0)
        {
            m--;
            i += d;
        }
  
        return m == 0 ? true : false;
    }
  
    // Function to implement
    // binary search
    static int maximumDistance(int arr[], int n, int m)
    {
  
        int low = 1, high = n - 1;
        int ans = 0;
  
        while (low <= high)
        {
  
            int mid = (low + high) / 2;
  
            // Check for valid distance i.e mid
            boolean flag = check(arr, n, m, mid);
  
            if (flag) 
            {
                ans = mid;
                low = mid + 1;
            } 
            else
            {
                high = mid - 1;
            }
        }
  
        return ans;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int n = 5, m = 3;
        int arr[] = new int[n];
  
        System.out.print(maximumDistance(arr, n, m));
  
    }
}
  
// This code is contributed by 29AjayKumar

Python

# Python3 program to Maximize distance between
# any two consecutive 1's after flipping M 0's
  
# Function to return the count
def check(arr, n, m, d):
      
    # Flipping zeros at distance "d"
    i = 0
    while (i < n and m > 0):
        m -= 1
        i += d
    if m == 0:
        return True
  
    return False
  
# Function to implement
# binary search
def maximumDistance(arr, n, m):
  
    low = 1
    high = n - 1
    ans = 0
  
    while (low <= high):
  
        mid = (low + high) // 2
  
        # Check for valid distance i.e mid
        flag = check(arr, n, m, mid)
  
        if (flag) :
            ans = mid
            low = mid + 1
        else :
            high = mid - 1
  
  
    return ans
  
# Driver code
  
n = 5
m = 3
arr = [0] * n
  
print(maximumDistance(arr, n, m))
  
# This code is contributed by mohit kumar 29

C#

// C# program to Maximize distance between
// any two consecutive 1's after flipping M 0's
using System;
  
class GFG 
{
  
    // Function to return the count
    static bool check(int []arr, int n, int m, int d)
    {
          
        // Flipping zeros at distance "d"
        int i = 0;
        while (i < n && m > 0)
        {
            m--;
            i += d;
        }
  
        return m == 0 ? true : false;
    }
  
    // Function to implement
    // binary search
    static int maximumDistance(int []arr, int n, int m)
    {
  
        int low = 1, high = n - 1;
        int ans = 0;
  
        while (low <= high)
        {
  
            int mid = (low + high) / 2;
  
            // Check for valid distance i.e mid
            bool flag = check(arr, n, m, mid);
  
            if (flag) 
            {
                ans = mid;
                low = mid + 1;
            } 
            else
            {
                high = mid - 1;
            }
        }
  
        return ans;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
  
        int n = 5, m = 3;
        int []arr = new int[n];
  
        Console.Write(maximumDistance(arr, n, m));
  
    }
}
  
// This code is contributed by PrinciRaj1992

输出：

2

时间复杂度： O(n * log(n))