README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1983.Widest%20Pair%20of%20Indices%20With%20Equal%20Range%20Sum/README_EN.md	Array Hash Table Prefix Sum

1983. Widest Pair of Indices With Equal Range Sum 🔒

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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

Solution 1: Prefix Sum + Hash Table

We observe that for any index pair $(i,j)$, if $nums1[i]+nums1[i+1]+...+nums1[j]=nums2[i]+nums2[i+1]+...+nums2[j]$, then $nums1[i]-nums2[i]+nums1[i+1]-nums2[i+1]+...+nums1[j]-nums2[j]=0$. If we subtract the corresponding elements of array $nums1$ and array $nums2$ to get a new array $nums$, the problem is transformed into finding the longest subarray in $nums$ such that the sum of the subarray is $0$. This can be solved using the prefix sum + hash table method.

We define a variable $s$ to represent the current prefix sum of $nums$, and use a hash table $d$ to store the first occurrence position of each prefix sum. Initially, $s=0$ and $d[0]=-1$.

Next, we traverse each element $x$ in the array $nums$, calculate the value of $s$, and then check whether $s$ exists in the hash table. If $s$ exists in the hash table, it means that there is a subarray $nums[d[s]+1,..i]$ such that the sum of the subarray is $0$, and we update the answer to $max(ans,i-d[s])$. Otherwise, we add the value of $s$ to the hash table, indicating that the first occurrence position of $s$ is $i$.

After the traversal, we can get the final answer.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}