# Longest subarray having count of 1s one more than count of 0s

Given an array of size **n** containing 0’s and 1’s only. The problem is to find the length of the longest subarray having count of 1’s one more than count of 0’s. **Examples:**

Input : arr = {0, 1, 1, 0, 0, 1} Output : 5 From index1to5. Input : arr[] = {1, 0, 0, 1, 0} Output : 1

**Approach:** Following are the steps:

- Consider all the 0’s in the array as ‘-1’.
- Initialize
**sum**= 0 and**maxLen**= 0. - Create a hash table having
**(sum, index)**tuples. - For i = 0 to n-1, perform the following steps:
- If arr[i] is ‘0’ accumulate ‘-1’ to
**sum**else accumulate ‘1’ to**sum**. - If sum == 1, update
**maxLen**= i+1. - Else check whether
**sum**is present in the hash table or not. If not present, then add it to the hash table as**(sum, i)**pair. - Check if
**(sum-1)**is present in the hash table or not. if present, then obtain index of**(sum-1)**from the hash table as**index**. Now check if maxLen is less than (i-index), then update**maxLen**= (i-index).

- If arr[i] is ‘0’ accumulate ‘-1’ to
- Return maxLen.

## C++

`// C++ implementation to find the length of` `// longest subarray having count of 1's one` `// more than count of 0's` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to find the length of longest` `// subarray having count of 1's one more` `// than count of 0's` `int` `lenOfLongSubarr(` `int` `arr[], ` `int` `n)` `{` ` ` `// unordered_map 'um' implemented as` ` ` `// hash table` ` ` `unordered_map<` `int` `, ` `int` `> um;` ` ` `int` `sum = 0, maxLen = 0;` ` ` `// traverse the given array` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// consider '0' as '-1'` ` ` `sum += arr[i] == 0 ? -1 : 1;` ` ` `// when subarray starts form index '0'` ` ` `if` `(sum == 1)` ` ` `maxLen = i + 1;` ` ` `// make an entry for 'sum' if it is` ` ` `// not present in 'um'` ` ` `else` `if` `(um.find(sum) == um.end())` ` ` `um[sum] = i;` ` ` `// check if 'sum-1' is present in 'um'` ` ` `// or not` ` ` `if` `(um.find(sum - 1) != um.end()) {` ` ` `// update maxLength` ` ` `if` `(maxLen < (i - um[sum - 1]))` ` ` `maxLen = i - um[sum - 1];` ` ` `}` ` ` `}` ` ` `// required maximum length` ` ` `return` `maxLen;` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `arr[] = { 0, 1, 1, 0, 0, 1 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << ` `"Length = "` ` ` `<< lenOfLongSubarr(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find the length of` `// longest subarray having count of 1's one` `// more than count of 0's` `import` `java.util.*;` `class` `GFG` `{` `// function to find the length of longest` `// subarray having count of 1's one more` `// than count of 0's` `static` `int` `lenOfLongSubarr(` `int` `arr[], ` `int` `n)` `{` ` ` `// unordered_map 'um' implemented as` ` ` `// hash table` ` ` `HashMap<Integer,` ` ` `Integer> um = ` `new` `HashMap<Integer,` ` ` `Integer>();` ` ` `int` `sum = ` `0` `, maxLen = ` `0` `;` ` ` `// traverse the given array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `// consider '0' as '-1'` ` ` `sum += arr[i] == ` `0` `? -` `1` `: ` `1` `;` ` ` `// when subarray starts form index '0'` ` ` `if` `(sum == ` `1` `)` ` ` `maxLen = i + ` `1` `;` ` ` `// make an entry for 'sum' if it is` ` ` `// not present in 'um'` ` ` `else` `if` `(!um.containsKey(sum))` ` ` `um. put(sum, i);` ` ` `// check if 'sum-1' is present in 'um'` ` ` `// or not` ` ` `if` `(um.containsKey(sum - ` `1` `))` ` ` `{` ` ` `// update maxLength` ` ` `if` `(maxLen < (i - um.get(sum - ` `1` `)))` ` ` `maxLen = i - um.get(sum - ` `1` `);` ` ` `}` ` ` `}` ` ` `// required maximum length` ` ` `return` `maxLen;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `1` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(` `"Length = "` `+` ` ` `lenOfLongSubarr(arr, n));` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python 3 implementation to find the length of` `# longest subarray having count of 1's one` `# more than count of 0's` `# function to find the length of longest` `# subarray having count of 1's one more` `# than count of 0's` `def` `lenOfLongSubarr(arr, n):` ` ` ` ` `# unordered_map 'um' implemented as` ` ` `# hash table` ` ` `um ` `=` `{i:` `0` `for` `i ` `in` `range` `(` `10` `)}` ` ` `sum` `=` `0` ` ` `maxLen ` `=` `0` ` ` `# traverse the given array` ` ` `for` `i ` `in` `range` `(n):` ` ` ` ` `# consider '0' as '-1'` ` ` `if` `arr[i] ` `=` `=` `0` `:` ` ` `sum` `+` `=` `-` `1` ` ` `else` `:` ` ` `sum` `+` `=` `1` ` ` `# when subarray starts form index '0'` ` ` `if` `(` `sum` `=` `=` `1` `):` ` ` `maxLen ` `=` `i ` `+` `1` ` ` `# make an entry for 'sum' if it is` ` ` `# not present in 'um'` ` ` `elif` `(` `sum` `not` `in` `um):` ` ` `um[` `sum` `] ` `=` `i` ` ` `# check if 'sum-1' is present in 'um'` ` ` `# or not` ` ` `if` `((` `sum` `-` `1` `) ` `in` `um):` ` ` `# update maxLength` ` ` `if` `(maxLen < (i ` `-` `um[` `sum` `-` `1` `])):` ` ` `maxLen ` `=` `i ` `-` `um[` `sum` `-` `1` `]` ` ` `# required maximum length` ` ` `return` `maxLen` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `1` `]` ` ` `n ` `=` `len` `(arr)` ` ` `print` `(` `"Length ="` `,lenOfLongSubarr(arr, n))` ` ` `# This code is contributed by` `# Surendra_Gangwar` |

## C#

`// C# implementation to find the length of` `// longest subarray having count of 1's one` `// more than count of 0's` `using` `System;` `using` `System.Collections.Generic;` ` ` `class` `GFG` `{` `// function to find the length of longest` `// subarray having count of 1's one more` `// than count of 0's` `static` `int` `lenOfLongSubarr(` `int` `[]arr, ` `int` `n)` `{` ` ` `// unordered_map 'um' implemented as` ` ` `// hash table` ` ` `Dictionary<` `int` `,` ` ` `int` `> um = ` `new` `Dictionary<` `int` `,` ` ` `int` `>();` ` ` `int` `sum = 0, maxLen = 0;` ` ` `// traverse the given array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `// consider '0' as '-1'` ` ` `sum += arr[i] == 0 ? -1 : 1;` ` ` `// when subarray starts form index '0'` ` ` `if` `(sum == 1)` ` ` `maxLen = i + 1;` ` ` `// make an entry for 'sum' if it is` ` ` `// not present in 'um'` ` ` `else` `if` `(!um.ContainsKey(sum))` ` ` `um.Add(sum, i);` ` ` `// check if 'sum-1' is present in 'um'` ` ` `// or not` ` ` `if` `(um.ContainsKey(sum - 1))` ` ` `{` ` ` `// update maxLength` ` ` `if` `(maxLen < (i - um[sum - 1]))` ` ` `maxLen = i - um[sum - 1];` ` ` `}` ` ` `}` ` ` `// required maximum length` ` ` `return` `maxLen;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]arr = { 0, 1, 1, 0, 0, 1 };` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(` `"Length = "` `+` ` ` `lenOfLongSubarr(arr, n));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` `// Javascript implementation to find the length of` `// longest subarray having count of 1's one` `// more than count of 0's` `// function to find the length of longest` `// subarray having count of 1's one more` `// than count of 0's` `function` `lenOfLongSubarr(arr, n)` `{` ` ` `// unordered_map 'um' implemented as` ` ` `// hash table` ` ` `var` `um = ` `new` `Map();` ` ` `var` `sum = 0, maxLen = 0;` ` ` `// traverse the given array` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `// consider '0' as '-1'` ` ` `sum += arr[i] == 0 ? -1 : 1;` ` ` `// when subarray starts form index '0'` ` ` `if` `(sum == 1)` ` ` `maxLen = i + 1;` ` ` `// make an entry for 'sum' if it is` ` ` `// not present in 'um'` ` ` `else` `if` `(!um.has(sum))` ` ` `um.set(sum, i);` ` ` `// check if 'sum-1' is present in 'um'` ` ` `// or not` ` ` `if` `(um.has(sum - 1)) {` ` ` `// update maxLength` ` ` `if` `(maxLen < (i - um.get(sum - 1)))` ` ` `maxLen = i - um.get(sum - 1);` ` ` `}` ` ` `}` ` ` `// required maximum length` ` ` `return` `maxLen;` `}` `// Driver program to test above` `var` `arr = [0, 1, 1, 0, 0, 1];` `var` `n = arr.length;` `document.write( ` `"Length = "` ` ` `+ lenOfLongSubarr(arr, n));` `// This code is contributed by itsok.` `</script>` |

**Output:**

Length = 5

**Time Complexity: **O(n) **Auxiliary Space:** O(n)

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