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.


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

Example 1


nums = [8, 7, 10, 11]




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;
        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)
        return ans;

                        Solution in Python : 
class Solution:
    def solve(self, nums):
            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)

View More Similar Problems

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →

Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

View Solution →