Minimum Value to Get Positive Step by Step Sum solution and explanation
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The problem
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] <= 100
Approach overview
Problem Overview: You are given an integer array nums. Starting with an unknown positive value startValue, you add each element in order and keep a running sum. The requirement is simple: the running sum must never drop below 1. The task is to compute the smallest possible starting value that guarantees this condition.
Approach 1: Calculate Minimum Start Value using Cumulative Sum (O(n) time, O(1) space)
Traverse the array while maintaining a running cumulative sum. Track the minimum value that this cumulative sum reaches during the traversal. If the lowest prefix sum becomes negative, the starting value must offset that dip so the total never falls below 1. Mathematically, the required start value is 1 - minPrefixSum. If the minimum prefix sum is already positive, the answer is simply 1. This works because every running total equals startValue + prefixSum, so ensuring the smallest prefix keeps the total ≥ 1 guarantees the rest will also be valid. The algorithm scans the array once and uses constant extra memory.
This approach relies on tracking prefix accumulation, which is a common technique in array problems and especially those involving prefix sum reasoning. You only store the current running sum and the minimum value encountered.
Approach 2: Prefix Sum with Constant Adjustment (O(n) time, O(1) space)
Another way to think about the problem is to explicitly compute the prefix sums and adjust them so the smallest prefix becomes 1. Iterate through nums, accumulate a prefix sum, and track the minimum prefix encountered. Instead of adjusting the prefix array itself, compute a constant shift that raises the minimum prefix to 1. That constant shift is the starting value. This interpretation emphasizes the idea that every prefix sum can be translated upward by the same constant to satisfy the constraint.
The logic is nearly identical to the cumulative-sum approach but framed as a transformation of the prefix sequence. It helps build intuition for similar problems where you normalize prefix values or maintain constraints across running totals. Because only two variables are maintained (current prefix and minimum prefix), the space complexity remains constant.
Recommended for interviews: The cumulative sum solution is the expected answer. It demonstrates that you recognize the prefix-sum pattern and can derive the minimum offset required to keep the running total positive. Explaining the minimum-prefix insight clearly usually matters more than the code itself. A brute-force simulation with increasing start values shows basic understanding, but interviewers typically expect the single-pass O(n) prefix analysis.
Approach 1: Calculate Minimum Start Value using Cumulative Sum
The idea is to simulate the step by step summation of the start value with the elements of the array. We maintain a running sum and adjust the start value such that this sum never drops below 1. At each step, if the running sum is less than 1, we calculate and update the new minimum start value needed to keep the running sum at least 1.
def min_start_value(nums):
min_sum = 0
current_sum = 0
for num in nums:
current_sum += num
min_sum = min(min_sum, current_sum)
return 1 - min_sum
# Example usage:
nums = [-3, 2, -3, 4, 2]
print(min_start_value(nums)) # Output: 5public class Solution {
public int minStartValue(int[] nums) {
int minSum = 0, currentSum = 0;
for (int num : nums) {
currentSum += num;
if (currentSum < minSum) {
minSum = currentSum;
}
}
return 1 - minSum;
}
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
}
}#include <vector>
#include <algorithm>
using namespace std;
int minStartValue(vector<int>& nums) {
int min_sum = 0, current_sum = 0;
for (int num : nums) {
current_sum += num;
min_sum = min(min_sum, current_sum);
}
return 1 - min_sum;
}
int main() {
vector<int> nums = {-3, 2, -3, 4, 2};
cout << minStartValue(nums) << endl; // Output: 5
}function minStartValue(nums) {
let minSum = 0, currentSum = 0;
for (let num of nums) {
currentSum += num;
minSum = Math.min(minSum, currentSum);
}
return 1 - minSum;
}
// Example usage:
const nums = [-3, 2, -3, 4, 2];
console.log(minStartValue(nums)); // Output: 5#include <stdio.h>
int minStartValue(int* nums, int numsSize) {
int min_sum = 0, current_sum = 0;
for (int i = 0; i < numsSize; i++) {
current_sum += nums[i];
if (current_sum < min_sum) {
min_sum = current_sum;
}
}
return 1 - min_sum;
}
int main() {
int nums[] = {-3, 2, -3, 4, 2};
printf("%d\n", minStartValue(nums, 5)); // Output: 5
return 0;
}using System;
public class Solution {
public int MinStartValue(int[] nums) {
int minSum = 0, currentSum = 0;
foreach (int num in nums) {
currentSum += num;
minSum = Math.Min(minSum, currentSum);
}
return 1 - minSum;
}
public static void Main() {
Solution sol = new Solution();
int[] nums = {-3, 2, -3, 4, 2};
Console.WriteLine(sol.MinStartValue(nums)); // Output: 5
}
}In this Python function, we calculate the running sum of the elements in the list and track the minimum value. The start value is calculated to ensure the cumulative sum never goes below 1, by adjusting with 1 minus the minimum sum.
Time Complexity: O(n), where n is the length of nums.
Space Complexity: O(1), since no additional space is utilized.
Approach 2: Prefix Sum with Constant Adjustment
This approach is similar to the first but conceptualized through using prefix sums and adjusting the needed start value based on the lowest prefix sum reached. We cumulatively add each number and check if the prefix sum dips below a certain threshold, indicating the minimal adjustment needed for the start value.
def min_start_value(nums):
min_prefix_sum = 0
prefix_sum = 0
for num in nums:
prefix_sum += num
min_prefix_sum = min(min_prefix_sum, prefix_sum)
return 1 - min_prefix_sum
# Example usage:
nums = [-3, 2, -3, 4, 2]
print(min_start_value(nums)) # Output: 5public class Solution {
public int minStartValue(int[] nums) {
int minPrefixSum = 0, prefixSum = 0;
for (int num : nums) {
prefixSum += num;
minPrefixSum = Math.min(minPrefixSum, prefixSum);
}
return 1 - minPrefixSum;
}
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
}
}#include <vector>
#include <algorithm>
using namespace std;
int minStartValue(vector<int>& nums) {
int min_prefix_sum = 0, prefix_sum = 0;
for (int num : nums) {
prefix_sum += num;
min_prefix_sum = min(min_prefix_sum, prefix_sum);
}
return 1 - min_prefix_sum;
}
int main() {
vector<int> nums = {-3, 2, -3, 4, 2};
cout << minStartValue(nums) << endl; // Output: 5
}function minStartValue(nums) {
let minPrefixSum = 0, prefixSum = 0;
for (let num of nums) {
prefixSum += num;
minPrefixSum = Math.min(minPrefixSum, prefixSum);
}
return 1 - minPrefixSum;
}
// Example usage:
const nums = [-3, 2, -3, 4, 2];
console.log(minStartValue(nums)); // Output: 5#include <stdio.h>
int minStartValue(int* nums, int numsSize) {
int min_prefix_sum = 0, prefix_sum = 0;
for (int i = 0; i < numsSize; i++) {
prefix_sum += nums[i];
if (prefix_sum < min_prefix_sum) {
min_prefix_sum = prefix_sum;
}
}
return 1 - min_prefix_sum;
}
int main() {
int nums[] = {-3, 2, -3, 4, 2};
printf("%d\n", minStartValue(nums, 5)); // Output: 5
return 0;
}using System;
public class Solution {
public int MinStartValue(int[] nums) {
int minPrefixSum = 0, prefixSum = 0;
foreach (int num in nums) {
prefixSum += num;
minPrefixSum = Math.Min(minPrefixSum, prefixSum);
}
return 1 - minPrefixSum;
}
public static void Main() {
Solution sol = new Solution();
int[] nums = {-3, 2, -3, 4, 2};
Console.WriteLine(sol.MinStartValue(nums)); // Output: 5
}
}The prefix sum is maintained over the iteration, comparing against a minimum threshold. The necessary start value ensures this cumulative prefix never dips below one during any point.
Time Complexity: O(n).
Space Complexity: O(1).
Approach 3: Default Approach
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);
}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
}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)
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Calculate Minimum Start Value using Cumulative Sum | Time Complexity: O(n), where n is the length of nums. |
| Prefix Sum with Constant Adjustment | Time Complexity: O(n). |
| Default Approach | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Cumulative Sum with Minimum Prefix Tracking | O(n) | O(1) | Best general solution. Single pass and minimal memory. |
| Prefix Sum with Constant Adjustment | O(n) | O(1) | Useful when reasoning about shifting prefix sums or teaching prefix-sum transformations. |
Frequently asked questions
Is Minimum Value to Get Positive Step by Step Sum easy or hard?
Minimum Value to Get Positive Step by Step Sum is classified as an Easy problem. The key insight is recognizing that the smallest prefix sum determines the minimum starting value needed to keep all running totals positive.
Minimum Value to Get Positive Step by Step Sum Python/Java solution
Both Python and Java implementations follow the same logic: iterate through the array, accumulate the running sum, track the minimum prefix value, and return max(1, 1 - minPrefix). The algorithm runs in O(n) time and O(1) space in both languages.
How to solve Minimum Value to Get Positive Step by Step Sum in O(n)?
Iterate through the array while computing a running prefix sum. Track the smallest prefix value encountered during the traversal. If the minimum prefix sum is negative, the required starting value is 1 minus that value; otherwise it is 1. This single pass guarantees the step-by-step total never drops below 1.
What is the best approach for Minimum Value to Get Positive Step by Step Sum?
The optimal approach tracks the running prefix sum and records the minimum value reached during traversal. The smallest starting value required is 1 minus that minimum prefix sum. This guarantees every step stays at least 1. The algorithm runs in O(n) time with O(1) extra space.
Is Minimum Value to Get Positive Step by Step Sum asked at Google/Amazon/Meta?
Prefix sum and running-total constraint problems like this frequently appear in interviews at companies such as Amazon, Google, and Meta. While this exact question may vary, the underlying technique of tracking minimum prefix sums is a common interview pattern.
What data structure is used in Minimum Value to Get Positive Step by Step Sum?
The solution primarily uses simple variables to maintain a running sum and the minimum prefix value. Conceptually it relies on the prefix sum technique applied to an array, but no additional data structures like stacks or hash maps are required.
What is the time complexity of Minimum Value to Get Positive Step by Step Sum?
The optimal solution runs in O(n) time because the array is scanned once while maintaining a running sum and the minimum prefix value. Only constant extra variables are used, so the space complexity is O(1).