给定一个大小为n的二进制数组arr[]和一个值k 。任务是找到最长子数组的长度，其 1 和 0 的计数差异等于k 。根据k的值，子数组中 1 的计数应等于或大于 0 的计数。
例子：
Input: arr[] = {0, 1, 1, 0, 1}, k = 2
Output: 4
The highlighted portion is the required subarray
{0, 1, 1, 0, 1}. In the subarray count of 1's is 3
and count of 0's is 1.
Therefore, difference in count = 3 - 1 = 2.
Input: arr[] = {1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1}, k = 0
Output: 6
The highlighted portion is the required subarray
{1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1}. In the subarray
count of 1's is 3 and count of 0's is 3.
Therefore, difference in count = 3 - 3 = 0.朴素的方法：考虑所有子数组的 1 和 0 计数的差异，并返回所需差异等于 ‘k’ 的最长子数组的长度。时间复杂度为 O(n^2)。
有效的方法：这个问题是寻找具有总和为 k 的最长子数组的变体。将arr[]中的所有 0 替换为 -1，然后找到总和等于 ‘k’的最长子数组 ‘arr’。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// function to find the length of the longest
// subarray having difference in the count
// of 1's and 0's equal to k
int lenOfLongSubarr(int arr[], int n, int k)
{
// unordered_map 'um' implemented
// as hash table
unordered_map um;
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++) {
// accumulate sum
sum += ((arr[i] == 0) ? -1 : arr[i]);
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
// required maximum length
return maxLen;
}
// Driver Code
int main()
{
int arr[] = { 0, 1, 1, 0, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
cout << "Length = "
<< lenOfLongSubarr(arr, n, k);
return 0;
} Java
// Java implementation of the above approach.
import java.util.HashMap;
import java.util.Map;
class GfG
{
// Function to find the length of the longest
// subarray having difference in the count
// of 1's and 0's equal to k
static int lenOfLongSubarr(int arr[], int n, int k)
{
// unordered_map 'um' implemented
// as hash table
HashMap um = new HashMap<>();
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++)
{
// accumulate sum
sum += ((arr[i] == 0) ? -1 : arr[i]);
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if
// it is not present in 'um'
if (!um.containsKey(sum))
um.put(sum, i);
// check if 'sum-k' is present
// in 'um' or not
if (um.containsKey(sum - k))
{
// update maxLength
if (maxLen < (i - um.get(sum - k)))
maxLen = i - um.get(sum - k);
}
}
// required maximum length
return maxLen;
}
// Driver code
public static void main(String []args)
{
int arr[] = { 0, 1, 1, 0, 1 };
int n = arr.length;
int k = 2;
System.out.println("Length = " + lenOfLongSubarr(arr, n, k));
}
}
// This code is contributed by Rituraj Jain Python
# Python3 implementation of above approach
# function to find the length of the longest
# subarray having difference in the count
# of 1's and 0's equal to k
def lenOfLongSubarr(arr, n, k):
# unordered_map 'um' implemented
# as hash table
um = dict()
Sum, maxLen = 0, 0
# traverse the given array
for i in range(n):
# accumulate sum
if arr[i] == 0:
Sum += -1
else:
Sum+=arr[i]
# when subarray starts from index '0'
if (Sum == k):
maxLen = i + 1
# make an entry for 'Sum' if it is
# not present in 'um'
if (Sum not in um.keys()):
um[Sum] = i
# check if 'Sum-k' is present in 'um'
# or not
if ((Sum - k) in um.keys()):
# update maxLength
if (maxLen < (i - um[Sum - k])):
maxLen = i - um[Sum - k]
# required maximum length
return maxLen
# Driver Code
arr = [0, 1, 1, 0, 1]
n = len(arr)
k = 2
print("Length = ",lenOfLongSubarr(arr, n, k))
# This code is contributed by mohit kumarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the length of the longest
// subarray having difference in the count
// of 1's and 0's equal to k
static int lenOfLongSubarr(int []arr,
int n, int k)
{
// unordered_map 'um' implemented
// as hash table
Dictionary um = new Dictionary();
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++)
{
// accumulate sum
sum += ((arr[i] == 0) ? -1 : arr[i]);
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if
// it is not present in 'um'
if (!um.ContainsKey(sum))
um.Add(sum, i);
// check if 'sum-k' is present
// in 'um' or not
if (um.ContainsKey(sum - k))
{
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
// required maximum length
return maxLen;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 0, 1, 1, 0, 1 };
int n = arr.Length;
int k = 2;
Console.WriteLine("Length = " +
lenOfLongSubarr(arr, n, k));
}
}
// This code is contributed by Princi Singh Javascript
输出：
Length = 4时间复杂度： O(n)
辅助空间： O(n)
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