# 1983. Widest Pair of Indices With Equal Range Sum

## Description

You are given two 0-indexed binary arrays nums1 and nums2. Find the widest pair of indices (i, j) such that i <= j and nums1[i] + nums1[i+1] + ... + nums1[j] == nums2[i] + nums2[i+1] + ... + nums2[j].

The widest pair of indices is the pair with the largest distance between i and j. The distance between a pair of indices is defined as j - i + 1.

Return the distance of the widest pair of indices. If no pair of indices meets the conditions, return 0.

Example 1:

Input: nums1 = [1,1,0,1], nums2 = [0,1,1,0]
Output: 3
Explanation:
If i = 1 and j = 3:
nums1[1] + nums1[2] + nums1[3] = 1 + 0 + 1 = 2.
nums2[1] + nums2[2] + nums2[3] = 1 + 1 + 0 = 2.
The distance between i and j is j - i + 1 = 3 - 1 + 1 = 3.


Example 2:

Input: nums1 = [0,1], nums2 = [1,1]
Output: 1
Explanation:
If i = 1 and j = 1:
nums1[1] = 1.
nums2[1] = 1.
The distance between i and j is j - i + 1 = 1 - 1 + 1 = 1.


Example 3:

Input: nums1 = [0], nums2 = [1]
Output: 0
Explanation:
There are no pairs of indices that meet the requirements.


Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• nums1[i] is either 0 or 1.
• nums2[i] is either 0 or 1.

## Solutions

• class Solution {
public int widestPairOfIndices(int[] nums1, int[] nums2) {
Map<Integer, Integer> d = new HashMap<>();
d.put(0, -1);
int n = nums1.length;
int s = 0;
int ans = 0;
for (int i = 0; i < n; ++i) {
s += nums1[i] - nums2[i];
if (d.containsKey(s)) {
ans = Math.max(ans, i - d.get(s));
} else {
d.put(s, i);
}
}
return ans;
}
}

• class Solution {
public:
int widestPairOfIndices(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> d;
d[0] = -1;
int ans = 0, s = 0;
int n = nums1.size();
for (int i = 0; i < n; ++i) {
s += nums1[i] - nums2[i];
if (d.count(s)) {
ans = max(ans, i - d[s]);
} else {
d[s] = i;
}
}
return ans;
}
};

• class Solution:
def widestPairOfIndices(self, nums1: List[int], nums2: List[int]) -> int:
d = {0: -1}
ans = s = 0
for i, (a, b) in enumerate(zip(nums1, nums2)):
s += a - b
if s in d:
ans = max(ans, i - d[s])
else:
d[s] = i
return ans


• func widestPairOfIndices(nums1 []int, nums2 []int) (ans int) {
d := map[int]int{0: -1}
s := 0
for i := range nums1 {
s += nums1[i] - nums2[i]
if j, ok := d[s]; ok {
ans = max(ans, i-j)
} else {
d[s] = i
}
}
return
}

• function widestPairOfIndices(nums1: number[], nums2: number[]): number {
const d: Map<number, number> = new Map();
d.set(0, -1);
const n: number = nums1.length;
let s: number = 0;
let ans: number = 0;
for (let i = 0; i < n; ++i) {
s += nums1[i] - nums2[i];
if (d.has(s)) {
ans = Math.max(ans, i - (d.get(s) as number));
} else {
d.set(s, i);
}
}
return ans;
}