Today I have done leetcode's March LeetCoding Challenge with cpp.

March LeetCoding Challenge 27

Description

Minimum Index of a Valid Split

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(vector<int>& nums) {
    map<int, int> count;
    int dominant = -1;
    for(auto n : nums) {
      count[n] += 1;
      if(count[n] > nums.size() / 2) dominant = n;
    }

    int allDominants = count[dominant];
    int leftCount = 0;
    for(int i = 0; i < nums.size(); ++i) {
      leftCount += (nums[i] == dominant);

      if(leftCount > (i + 1) / 2 && (allDominants - leftCount) > (nums.size() - i - 1) / 2) return i;
    }

    return -1;
  }
};

// Accepted
// 917/917 cases passed (75 ms)
// Your runtime beats 10.66 % of cpp submissions
// Your memory usage beats 36.07 % of cpp submissions (99.3 MB)