2888. Minimum Index Of A Valid Split¶

Difficulty: Medium

2888. Minimum Index of a Valid Split

Medium

An element x of an integer array arr of length m is dominant if more than half the elements of arr have a value of x.

You are given a 0-indexed integer array nums of length n with one dominant element.

You can split nums at an index i into two arrays nums[0, ..., i] and nums[i + 1, ..., n - 1], but the split is only valid if:

0 <= i < n - 1
nums[0, ..., i], and nums[i + 1, ..., n - 1] have the same dominant element.

Here, nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j, both ends being inclusive. Particularly, if j < i then nums[i, ..., j] denotes an empty subarray.

Return the minimum index of a valid split. If no valid split exists, return -1.

Example 1:

Input: nums = [1,2,2,2]
Output: 2
Explanation: We can split the array at index 2 to obtain arrays [1,2,2] and [2]. 
In array [1,2,2], element 2 is dominant since it occurs twice in the array and 2 * 2 > 3. 
In array [2], element 2 is dominant since it occurs once in the array and 1 * 2 > 1.
Both [1,2,2] and [2] have the same dominant element as nums, so this is a valid split. 
It can be shown that index 2 is the minimum index of a valid split.

Example 2:

Input: nums = [2,1,3,1,1,1,7,1,2,1]
Output: 4
Explanation: We can split the array at index 4 to obtain arrays [2,1,3,1,1] and [1,7,1,2,1].
In array [2,1,3,1,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
In array [1,7,1,2,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
Both [2,1,3,1,1] and [1,7,1,2,1] have the same dominant element as nums, so this is a valid split.
It can be shown that index 4 is the minimum index of a valid split.

Example 3:

Input: nums = [3,3,3,3,7,2,2]
Output: -1
Explanation: It can be shown that there is no valid split.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
nums has exactly one dominant element.

Solution¶

class Solution {
    public int minimumIndex(List<Integer> nums) {
        int n = nums.size();
        HashMap<Integer, Integer> map1 = new HashMap<>();
        HashMap<Integer, Integer> map2 = new HashMap<>();
        for (int i = 0; i < n; i++) map1.put(nums.get(i), map1.getOrDefault(nums.get(i), 0) + 1);
        for (int i = 0; i < n; i++) {
            int current = nums.get(i);
            map2.put(current, map2.getOrDefault(current, 0) + 1);
            map1.put(current, map1.getOrDefault(current, 0) - 1);
            if (map2.getOrDefault(current, 0) * 2 > i + 1 && map1.getOrDefault(current, 0) * 2 > n - i - 1) return i;
        }
        return -1;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here