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Leetcode 2996. Smallest Missing Integer Greater Than Sequential Prefix Sum

Problem Statement

You are given a zero-based integer array arr containing n positive integers. You need to find the smallest positive integer greater than the sum of the elements in the prefix up to some index. More formally, you need to find the smallest integer k such that k > arr[0] + arr[1] + ... + arr[j] for any 0 <= j < n.

Example:

Input: arr = [1, 2, 5, 10]
Output: 19

Clarifying Questions

  1. Constraints on Array Elements and Size:
    • What is the range of values for the elements in arr?
    • What is the maximum size n of the array?
  2. Unique Elements:
    • Can the array contain duplicate values?
  3. Input Edge Cases:
    • What should be the output if the input array is empty?
    • How should negative numbers be handled if they appear in the array?

Code

#include <iostream>
#include <vector>

int findSmallestMissingIntegerGreaterThanPrefixSum(const std::vector<int>& arr) {
    int currentSum = 0;
    for (int num : arr) {
        currentSum += num;
    }
    return currentSum + 1;
}

int main() {
    std::vector<int> arr = {1, 2, 5, 10};
    std::cout << "Smallest Missing Integer Greater Than Prefix Sum: " 
              << findSmallestMissingIntegerGreaterThanPrefixSum(arr) << std::endl;
    return 0;
}

Strategy

The provided example and the described problem allow us to devise a straightforward approach:

  1. Prefix Sum Calculation:
    • Calculate the total sum of all elements in the array. Let’s call this totalSum.
  2. Smallest Integer Greater Than Prefix Sum:
    • The desired smallest missing integer `k` would be totalSum + 1, because totalSum is the sum of all elements, and any integer > totalSum that starts only after totalSum + 1.

Steps:

Time Complexity

The solution involves a single scan through the array to calculate the sum, which gives it a:

This approach ensures efficiency and simplicity, taking advantage of basic arithmetic operations and a single traversal through the input list.

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