Minimum length subarray of 1s in a Binary Array

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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 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; // intitialize 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);        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; // intitialize 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; # intitialize 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; // intitialize 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..

Output:

2

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

My Personal Notes arrow_drop_up Striver(underscore)79 at Codechef and codeforces D

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Improved By : jit_t, Rajput-Ji

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