41. First Missing Positive

Explanation

To solve this problem in O(n) time and O(1) auxiliary space, we can utilize the array itself to keep track of the missing positive integers. We iterate through the array and place each element at its correct position (i.e., nums[i] should be placed at index nums[i] - 1). After the rearrangement, we then iterate through the array again to find the first index where the value does not match the index (indicating a missing positive integer). If no such index is found, then the missing integer is one more than the length of the array.

Time Complexity: O(n)
Space Complexity: O(1)

View Leetcode Problem

class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) {
                int temp = nums[nums[i] - 1];
                nums[nums[i] - 1] = nums[i];
                nums[i] = temp;
            }
        }
        
        for (int i = 0; i < n; i++) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }
        
        return n + 1;
    }
}

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