Win After Last Round - Microsoft Top Interview Questions


Problem Statement :


You are given a list of integers nums of length n representing the current score of swimmers in a competition. 

There is one more round to swim and the first place winner for this round gets n points, second place n-1 points, etc. and the last place gets 1 point.

Return the number of swimmers that can still win the competition after the last round. If you tie for first in points, this still counts as winning.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [8, 7, 10, 11]

Output

3

Explanation

The swimmers that currently have 8, 10 and 11 points can all win if final score is [12, 10, 12, 12]. That is, the 8 point swimmer gets first place, 7 point swimmer swimmer gets second, 10 point swimmer 
gets third, and 11 point swimmer gets last place.


Even if the 7 point swimmer gets first place and has final score of 11 points, 8 point swimmer gets 
second, the third place person would still get 2 points so the last two swimmers would still get at least 12 points. So the 7 point swimmer cannot win the competition.



Solution :



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                        Solution in C++ :

int solve(vector<int>& nums) {
    // Protect vs empty vector
    if (nums.empty()) return 0;

    // Sort the array to make the rest of the code easier.
    sort(nums.begin(), nums.end());
    const int n = nums.size();

    // We need to find which final score will give us the winning score. This is done by adding the
    // smallest possible points to the best previous scores. We start by adding 1 point to the best
    // score.
    int best = nums[n - 1] + 1;
    // Keep adding more points to swimmers from the best to the worst and check if they have the new
    // max score.
    for (int i = n - 2; i >= 0; --i) {
        best = max<int>(best, nums[i] + (n - i));
    }

    // All we need to do now is to go from the smallest score, add max possible points and check if
    // it beats the winning score.
    for (int i = 0; i < n; ++i) {
        if (nums[i] + n >= best) return n - i;
    }

    return 1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        int N = nums.length;
        Arrays.sort(nums);
        int best = 0; // minimum score of the winner
        for (int i = 0; i < N; i++) best = Math.max(best, nums[i] + (N - i));
        int ans = 0;
        for (int i = 0; i < N; i++) {
            if (nums[i] + N >= best)
                ans++;
        }
        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        nums.sort(reverse=True)
        try:
            cut = max(num + i for i, num in enumerate(nums, 1))
        except ValueError:
            return 0
        for i, num in enumerate(nums):
            if num + len(nums) < cut:
                return i
        return len(nums)
                    


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