Question

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 $nums$.

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

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;
  }
};

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;
}