**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 :

` ````
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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