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Leetcode 1413. Minimum Value to Get Positive Step by Step Sum

Problem Statement

You are given an integer array nums. You start with an initial value startValue = 0.

In each step, you are required to add the current element of the array to startValue.

Return the minimum positive value of startValue such that the step by step sum is always positive.

Clarifying Questions

  1. Range of Input Values: What are the constraints on the values of nums and its length?
    • nums can be of length between 1 and 100.
    • Each element in nums can be in the range of -100 to 100.
  2. Output: Should I return the minimum positive integer as the output?
    • Yes, you need to return the smallest positive integer value for startValue such that the cumulative sum never drops to zero or below during the iteration through the array.

Strategy

  1. Initialization: Initialize minVal to a high positive value (e.g., Integer.MAX_VALUE) to keep track of the minimum step-by-step sum encountered during the iteration.
  2. Iterate and Track: Iterate through the array while maintaining a running sum and update minVal with the smallest running sum.
  3. Determine Start Value: The minimum positive startValue will be 1 - minVal if minVal is negative or zero. If minVal is positive, startValue will simply be 1.

Code

public class Solution {
    public int minStartValue(int[] nums) {
        int minVal = Integer.MAX_VALUE;
        int runningSum = 0;
        for (int num : nums) {
            runningSum += num;
            minVal = Math.min(minVal, runningSum);
        }
        
        // If minVal is negative or zero, the minimum start value to get positive sums is 1 - minVal
        // If minVal is positive, the minimum start value is 1
        return minVal < 1 ? 1 - minVal : 1;
    }
    
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] nums = {-3, 2, -3, 4, 2};
        System.out.println(sol.minStartValue(nums)); // Output: 5
    }
}

Time Complexity

This strategy ensures that the solution is efficient and meets the problem’s constraints.

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