Minimum length subarray of 1s in a Binary Array

Last Updated : 28 Mar, 2023

Given binary array. The task is to find the length of subarray with minimum number of 1s.
Note: It is guaranteed that there is atleast one 1 present in the array.
Examples :

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

Simple Solution: A simple solution is to consider every subarray and count 1’s in every subarray. Finally return return size of minimum length subarray of 1s.

C++

   #include <bits/stdc++.h>
using namespace std;
 
int subarrayWithMinOnes(int arr[], int n)
{
    int ans = INT_MAX;
 
    // consider all subarrays starting from index i
    for (int i = 0; i < n; i++) {
        // consider all subarrays ending at index j
        for (int j = i+1; j < n; j++) {
            int count = 0;
            bool flag = true;
            // count the number of 1s in the current subarray
            for (int k = i; k <= j; k++) {
                if (arr[k] != 1) {
                    flag = false;
                    break;
                }
                else
                    count++;
            }
            // if current subarray has all 1s, update ans
            if (flag)
                ans = min(ans, count);
        }
    }
    return ans;
}
 
int main()
{
    int arr[] = { 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << subarrayWithMinOnes(arr, n) << endl;
    return 0;
}
//This code is conntributed by Naveen Gujjar
 
  

Java

   import java.util.*;
 
public class Main {
    public static int subarrayWithMinOnes(int[] arr, int n) {
        int ans = Integer.MAX_VALUE;
 
        // consider all subarrays starting from index i
        for (int i = 0; i < n; i++) {
            // consider all subarrays ending at index j
            for (int j = i+1; j < n; j++) {
                int count = 0;
                boolean flag = true;
                // count the number of 1s in the current subarray
                for (int k = i; k <= j; k++) {
                    if (arr[k] != 1) {
                        flag = false;
                        break;
                    }
                    else
                        count++;
                }
                // if current subarray has all 1s, update ans
                if (flag)
                    ans = Math.min(ans, count);
            }
        }
        return ans;
    }
 
    public static void main(String[] args) {
        int[] arr = { 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1 };
        int n = arr.length;
        System.out.println(subarrayWithMinOnes(arr, n));
    }
}
 
  

Python3

   def subarrayWithMinOnes(arr, n):
    ans = float('inf')
 
    # consider all subarrays starting from index i
    for i in range(n):
       
        # consider all subarrays ending at index j
        for j in range(i+1, n):
            count = 0
            flag = True
             
            # count the number of 1s in the current subarray
            for k in range(i, j+1):
                if arr[k] != 1:
                    flag = False
                    break
                else:
                    count += 1
                     
            # if current subarray has all 1s, update ans
            if flag:
                ans = min(ans, count)
    return ans
 
arr = [1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1]
n = len(arr)
print(subarrayWithMinOnes(arr, n))
 
  

Javascript

   function subarrayWithMinOnes(arr, n) {
    let ans = Infinity;
    for (let i = 0; i < n; i++) {
        for (let j = i+1; j < n; j++) {
            let count = 0;
            let flag = true;
            for (let k = i; k <= j; k++) {
                if (arr[k] !== 1) {
                    flag = false;
                    break;
                } else {
                    count++;
                }
            }
            if (flag) {
                ans = Math.min(ans, count);
            }
        }
    }
    return ans;
}
 
let arr = [1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1];
let n = arr.length;
console.log(subarrayWithMinOnes(arr, n));
 
  

C#

   using System;
 
class Program {
    static int SubarrayWithMinOnes(int[] arr, int n)
    {
        int ans = int.MaxValue;
        // consider all subarrays starting from index i
        for (int i = 0; i < n; i++) {
            // consider all subarrays ending at index j
            for (int j = i + 1; j < n; j++) {
                int count = 0;
                bool flag = true;
                // count the number of 1s in the current
                // subarray
                for (int k = i; k <= j; k++) {
                    if (arr[k] != 1) {
                        flag = false;
                        break;
                    }
                    else
                        count++;
                }
                // if current subarray has all 1s, update
                // ans
                if (flag)
                    ans = Math.Min(ans, count);
            }
        }
        return ans;
    }
 
    static void Main(string[] args)
    {
        int[] arr = { 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1 };
        int n = arr.Length;
        Console.WriteLine(SubarrayWithMinOnes(arr, n));
    }
}
// This code is contributed by sarojmcy2e
 
  

Output

Time complexity: O(n^3)
Auxiliary Space: O(1)

Efficient Solution: An efficient solution is traverse array from left to right. If we see a 1, we increment count. If we see a 0, and count of 1s so far is positive, calculate minimum of count and result and reset count to zero.
Below is the implementation of the above approach:

C++

   // C++ program to count minimum length
// subarray of 1's in a binary array.
#include <bits/stdc++.h>
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

   <script>
 
// 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 
 
</script>
 
  

Output:

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

Minimum LCM of all subarrays of length at least 2

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Minimum length subarray of 1s in a Binary Array

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