Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
Given an array of integers nums, you start with an initial positive value startValue.
In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right).
Return the minimum positive value of startValue such that the step by step sum is never less than 1.
Example 1:
Input: nums = [-3,2,-3,4,2] Output: 5 Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1. step by step sum startValue = 4 | startValue = 5 | nums (4 -3 ) = 1 | (5 -3 ) = 2 | -3 (1 +2 ) = 3 | (2 +2 ) = 4 | 2 (3 -3 ) = 0 | (4 -3 ) = 1 | -3 (0 +4 ) = 4 | (1 +4 ) = 5 | 4 (4 +2 ) = 6 | (5 +2 ) = 7 | 2
Example 2:
Input: nums = [1,2] Output: 1 Explanation: Minimum start value should be positive.
Example 3:
Input: nums = [1,-2,-3] Output: 5
Constraints:
1 <= nums.length <= 100-100 <= nums[i] <= 100Problem summary: Given an array of integers nums, you start with an initial positive value startValue. In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right). Return the minimum positive value of startValue such that the step by step sum is never less than 1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[-3,2,-3,4,2]
[1,2]
[1,-2,-3]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1413: Minimum Value to Get Positive Step by Step Sum
class Solution {
public int minStartValue(int[] nums) {
int s = 0;
int t = Integer.MAX_VALUE;
for (int num : nums) {
s += num;
t = Math.min(t, s);
}
return Math.max(1, 1 - t);
}
}
// Accepted solution for LeetCode #1413: Minimum Value to Get Positive Step by Step Sum
func minStartValue(nums []int) int {
s, t := 0, 10000
for _, num := range nums {
s += num
if s < t {
t = s
}
}
if t < 0 {
return 1 - t
}
return 1
}
# Accepted solution for LeetCode #1413: Minimum Value to Get Positive Step by Step Sum
class Solution:
def minStartValue(self, nums: List[int]) -> int:
s, t = 0, inf
for num in nums:
s += num
t = min(t, s)
return max(1, 1 - t)
// Accepted solution for LeetCode #1413: Minimum Value to Get Positive Step by Step Sum
impl Solution {
pub fn min_start_value(nums: Vec<i32>) -> i32 {
let mut sum = 0;
let mut min = i32::MAX;
for num in nums.iter() {
sum += num;
min = min.min(sum);
}
(1).max(1 - min)
}
}
// Accepted solution for LeetCode #1413: Minimum Value to Get Positive Step by Step Sum
function minStartValue(nums: number[]): number {
let sum = 0;
let min = Infinity;
for (const num of nums) {
sum += num;
min = Math.min(min, sum);
}
return Math.max(1, 1 - min);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.