287_Find_duplicate_number

// Leetcode 287: Find the Duplicate Number
// Detailed Explanation is available here: https://youtu.be/GOYMcrvg-Ck

// C++

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int n = nums.size();
        int idx = 0;
        int curr_max = 0;
        for(int i = 0; i < n; ++i){
            int id = nums[i] % n;
            nums[id] += n;
        }
        for(int i = 0; i < n; ++i){
            if(nums[i] > curr_max){
                curr_max = nums[i];
                idx = i;
            }
            nums[i] %= n;
        }
        return idx;
    }
};

// Java:
class Solution {
    public int findDuplicate(int[] nums) {
        int n = nums.length;
        int idx = 0;
        int curr_max = 0;
        for(int i = 0; i < n; ++i){
            int id = nums[i] % n;
            nums[id] += n;
        }
        for(int i = 0; i < n; ++i){
            if(nums[i] > curr_max){
                curr_max = nums[i];
                idx = i;
            }
            nums[i] %= n;
        }
        return idx;
    }
}

# Python3:
class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        n = len(nums)
        idx = 0
        curr_max = 0
        for i in range(n):
            id = nums[i] % n
            nums[id] += n
        
        for i in range(n):
            if nums[i] > curr_max:
                curr_max = nums[i]
                idx = i
            nums[i] %= n
        
        return idx
        
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