Minimum length subarray of 1s in a Binary Array
Last Updated :
28 Mar, 2023
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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.
#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
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 Gujjarimport 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));
}
}
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))
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));
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
2
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++ 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;
}
// C++ program to count minimum length// subarray of 1's in a binary array.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 codeint 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 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.
# 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# 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..
<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:
2
Time Complexity: O(N)
Auxiliary Space: O(1)
