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Leetcode 2996. Smallest Missing Integer Greater Than Sequential Prefix Sum

Problem Statement

Given an array of positive integers nums, find the smallest missing positive integer greater than the sum of the first i elements for each prefix sum. The result should be the smallest integer that cannot be formed by the sum of the prefix elements plus these integers.

Clarifying Questions

  1. What should the function return if the array is empty?
    • If the array is empty, the smallest missing positive integer would be 1.
  2. Can there be duplicate elements in the array?
    • Yes, there can be duplicate elements in the array.
  3. Is there any upper limit on the size of the array or the values within it?
    • For practical purposes, we can assume the array can be large, and values are positive integers within reasonable computational limits.

Strategy

  1. Prefix Sum Calculation:
    • Calculate the prefix sum as we iterate through the array. This step doesn’t require an auxiliary list; we can maintain a running total.
  2. Finding the Smallest Missing Integer:
    • The key observation is that if we can form every integer from 1 to the current prefix sum, the next integer we can’t form with the given prefix is the prefix sum plus 1.
    • Iterate through each element, keeping track of the prefix sum.
    • At each step, check if the current element can help form new integers up to the current prefix sum.
  3. Edge Cases:
    • If the array is empty, we should return 1.
    • If the smallest missing positive integer is larger than any prefix sum found by the end of the array.

Code

public class SmallestMissingInteger {
    public int findSmallestMissing(int[] nums) {
        if (nums.length == 0) {
            return 1;
        }
        
        int prefixSum = 0;
        for (int num : nums) {
            if (num > prefixSum + 1) {
                break;
            }
            prefixSum += num;
        }
        
        return prefixSum + 1;
    }
}

// Example usage:
// SmallestMissingInteger solution = new SmallestMissingInteger();
// int result = solution.findSmallestMissing(new int[]{1, 1, 2, 5, 3});
// System.out.println(result);  // Output: 13

Time Complexity

This approach is efficient and leverages a simple scan through the array to solve the problem.

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