First Missing Positive - Amazon Top Interview Questions


Problem Statement :


Given a list of integers nums, find the first missing positive integer. In other words, find the lowest positive integer that does not exist in the list. The list can contain duplicates and negative numbers as well.

Constraints

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

Example 1

Input

nums = [1, 2, 3]

Output

4

Example 2

Input

nums = [3, 4, -1, 1]

Output

2

Example 3

Input

nums = [1, 2, 0]

Output

3

Example 4


Input

nums = [-1, -2, -3]

Output

1


Solution :



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                        Solution in C++ :

int solve(vector<int>& nums) {
    int N = nums.size();
    for (int i = 0; i < N;) {
        if (nums[i] > 0 && nums[i] != i + 1 && nums[i] <= N && nums[i] != nums[nums[i] - 1]) {
            swap(nums[i], nums[nums[i] - 1]);
        } else {
            i++;
        }
    }

    for (int i = 0; i < N; i++) {
        if (nums[i] != i + 1) return i + 1;
    }
    return N + 1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        boolean[] exists = new boolean[nums.length];

        for (Integer i : nums) {
            if (i > 0 && i <= nums.length)
                exists[i - 1] = true;
        }

        for (int i = 0; i < exists.length; i++) {
            if (exists[i] == false)
                return i + 1;
        }

        return nums.length + 1;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        n = len(nums)
        for i in range(n):
            correctPos = nums[i] - 1  # 0-indexng
            while 1 <= nums[i] <= n and nums[i] != nums[correctPos]:
                nums[i], nums[correctPos] = nums[correctPos], nums[i]
                # nums[i] has changed
                correctPos = nums[i] - 1

        for i in range(n):
            if i + 1 != nums[i]:
                return i + 1
        return n + 1


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