二进制数组中 1 的最小长度子数组

给定二进制数组。任务是找到最小数量为 1 的子数组的长度。
注意：保证数组中至少存在一个 1。
例子：

Input : arr[] = {1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1} 
Output : 3 
Minimum length subarray of 1s is {1, 1}.
Input : arr[] = {0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1} 
Output : 1 
 

简单的解决方案：一个简单的解决方案是考虑每个子数组并计算每个子数组中的 1。最后返回 1s 的最小长度子数组的返回大小。
有效的解决方案：一个有效的解决方案是从左到右遍历数组。如果我们看到 1，我们增加计数。如果我们看到 0，并且到目前为止 1 的计数是正数，则计算计数和结果的最小值并将计数重置为零。
下面是上述方法的实现：

C++

// C++ program to count minimum length
// subarray of 1's in a binary array.
#include 
using namespace std;
 
// Function to count minimum length subarray
// of 1's in binary array arr[0..n-1]
int getMinLength(bool arr[], int n)
{
    int count = 0; // initialize count
    int result = INT_MAX; // initialize result
 
    for (int i = 0; i < n; i++) {
        if (arr[i] == 1) {
            count++;
        }
        else {
            if (count != 0)
                result = min(result, count);
            count = 0;
        }
    }
 
    return result;
}
 
// Driver code
int main()
{
    bool arr[] = { 1, 1, 0, 0, 1, 1, 1, 0,
                   1, 1, 1, 1 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << getMinLength(arr, n) << endl;
 
    return 0;
}

Java

// Java program to count minimum length
// subarray of 1's in a binary array.
import java.io.*;
 
class GFG
{
     
// Function to count minimum length subarray
// of 1's in binary array arr[0..n-1]
static int getMinLength(double arr[], int n)
{
    int count = 0; // initialize count
    int result = Integer.MAX_VALUE; // initialize result
 
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == 1)
        {
            count++;
        }
        else
        {
            if (count != 0)
                result = Math.min(result, count);
            count = 0;
        }
    }
 
    return result;
}
 
// Driver code
public static void main (String[] args)
{
    double arr[] = { 1, 1, 0, 0, 1, 1, 1, 0,
                1, 1, 1, 1 };
    int n = arr.length;
    System.out.println (getMinLength(arr, n));
 
}
}
 
// This code is contributed by ajit.

Python3

# Python program to count minimum length
# subarray of 1's in a binary array.
import sys
 
# Function to count minimum length subarray
# of 1's in binary array arr[0..n-1]
def getMinLength(arr, n):
    count = 0; # initialize count
    result = sys.maxsize ; # initialize result
 
    for i in range(n):
        if (arr[i] == 1):
            count+=1;
        else:
            if(count != 0):
                result = min(result, count);
            count = 0;
 
    return result;
 
# Driver code
arr = [ 1, 1, 0, 0, 1, 1, 1, 0,
                1, 1, 1, 1 ];
 
n = len(arr);
 
print(getMinLength(arr, n));
 
# This code is contributed by Rajput-Ji

C#

// C# program to count minimum length
// subarray of 1's in a binary array.
using System;
 
class GFG
{
     
// Function to count minimum length subarray
// of 1's in binary array arr[0..n-1]
static int getMinLength(double []arr, int n)
{
    int count = 0; // initialize count
    int result = int.MaxValue; // initialize result
 
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == 1)
        {
            count++;
        }
        else
        {
            if (count != 0)
                result = Math.Min(result, count);
            count = 0;
        }
    }
 
    return result;
}
 
// Driver code
static public void Main ()
{
    double []arr = { 1, 1, 0, 0, 1, 1,
                     1, 0, 1, 1, 1, 1 };
    int n = arr.Length;
    Console.WriteLine(getMinLength(arr, n));
}
}
 
// This code is contributed by Tushil..

Javascript

 
// javascript program to count minimum length
// subarray of 1's in a binary array.
  
// Function to count minimum length subarray
// of 1's in binary array arr[0..n-1]
function getMinLength(arr, n)
{
// initialize count
    var count = 0;
// initialize result
    var result = Number.MAX_VALUE;
 
    for (i = 0; i < n; i++)
    {
        if (arr[i] == 1)
        {
            count++;
        }
        else
        {
            if (count != 0)
                result = Math.min(result, count);
            count = 0;
        }
    }
 
    return result;
}
 
// Driver code
var arr = [ 1, 1, 0, 0, 1, 1, 1, 0,
           1, 1, 1, 1 ];
var n = arr.length;
document.write(getMinLength(arr, n));
 
// This code is contributed by Amit Katiyar
 

输出：

2

时间复杂度：O(N)
辅助空间：O(1)