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1413. Minimum Value to Get Positive Step by Step Sum

EasyArrayPrefix Sum
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Problem Description

You are given an array of integers nums and need to find the minimum positive starting value such that when you calculate the cumulative sum (starting value + running sum of array elements), the result never drops below 1.

Starting with an initial positive value startValue, you iterate through the array from left to right, adding each element to your running total. At each step, your current sum must be at least 1.

For example, if nums = [-3, 2, -3, 4, 2] and startValue = 5:

  • Step 1: 5 + (-3) = 2 (≥ 1, valid)
  • Step 2: 2 + 2 = 4 (≥ 1, valid)
  • Step 3: 4 + (-3) = 1 (≥ 1, valid)
  • Step 4: 1 + 4 = 5 (≥ 1, valid)
  • Step 5: 5 + 2 = 7 (≥ 1, valid)

The solution tracks the minimum value reached during the cumulative sum calculation. If the minimum cumulative sum is t, then the starting value must be at least 1 - t to ensure the sum never drops below 1. The answer is max(1, 1 - t) since the starting value must be positive.

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Intuition

The key insight is that we need to find the "worst case" scenario - the point where our running sum reaches its lowest value. If we can ensure our sum stays positive at this critical point, it will remain positive everywhere else.

Think of it like walking along a path with hills and valleys. We start at some height (startValue) and each step either takes us up or down based on the array values. To never touch the ground (sum < 1), we need to start high enough to clear the deepest valley.

As we traverse the array calculating cumulative sums, we track the minimum value reached. This minimum represents our deepest valley. If this minimum is negative or zero, we need to lift our starting point high enough so that even at this lowest point, we're still at height 1 or above.

The mathematical relationship becomes clear: if our minimum cumulative sum is t, then starting at value x, our lowest point will be x + t. For this to be at least 1, we need x + t ≥ 1, which means x ≥ 1 - t.

Since we also need the starting value to be positive (at least 1), the answer is max(1, 1 - t). This elegantly handles both cases:

  • If the cumulative sum never goes negative (t ≥ 0), we can start at 1
  • If the cumulative sum does go negative (t < 0), we need to start at 1 - t to compensate for the drop

Learn more about Prefix Sum patterns.

Solution Approach

The implementation uses a single-pass algorithm with constant space complexity to find the minimum starting value.

We initialize two variables:

  • s = 0: Tracks the running cumulative sum as we iterate through the array
  • t = inf: Stores the minimum cumulative sum encountered so far (initialized to infinity)

The algorithm proceeds in these steps:

  1. Iterate through the array: For each element num in nums:

    • Add num to the running sum: s += num
    • Update the minimum cumulative sum: t = min(t, s)

  2. Calculate the minimum starting value: After finding the minimum cumulative sum t, compute the result as max(1, 1 - t)

Let's trace through an example with nums = [-3, 2, -3, 4, 2]:

  • Initially: s = 0, t = inf
  • After -3: s = -3, t = -3
  • After 2: s = -1, t = -3
  • After -3: s = -4, t = -4
  • After 4: s = 0, t = -4
  • After 2: s = 2, t = -4

The minimum cumulative sum is -4, so the minimum starting value is max(1, 1 - (-4)) = max(1, 5) = 5.

This approach is optimal because:

  • Time Complexity: O(n) - single pass through the array
  • Space Complexity: O(1) - only uses two variables regardless of input size
  • Correctness: By tracking the minimum point, we guarantee that if the sum is valid at the lowest point, it will be valid everywhere

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Example Walkthrough

Let's walk through a small example with nums = [1, -2, -3] to illustrate the solution approach.

Step 1: Track cumulative sums and find the minimum

We'll iterate through the array, maintaining a running sum s and tracking the minimum value t:

  • Start: s = 0, t = infinity
  • Process 1:
    • s = 0 + 1 = 1
    • t = min(infinity, 1) = 1
  • Process -2:
    • s = 1 + (-2) = -1
    • t = min(1, -1) = -1
  • Process -3:
    • s = -1 + (-3) = -4
    • t = min(-1, -4) = -4

Step 2: Calculate minimum starting value

Our minimum cumulative sum is t = -4. To ensure the sum never drops below 1, we need:

  • Starting value = max(1, 1 - t)
  • Starting value = max(1, 1 - (-4))
  • Starting value = max(1, 5)
  • Starting value = 5

Step 3: Verify the solution

Let's confirm that starting with 5 keeps our sum at least 1:

  • Start with 5
  • After 1: 5 + 1 = 6 ✓ (≥ 1)
  • After -2: 6 + (-2) = 4 ✓ (≥ 1)
  • After -3: 4 + (-3) = 1 ✓ (≥ 1)

The minimum point occurs at the end where we reach exactly 1. This confirms our starting value of 5 is correct and minimal.

Solution Implementation

1class Solution:
2    def minStartValue(self, nums: List[int]) -> int:
3        # Initialize cumulative sum and minimum cumulative sum
4        cumulative_sum = 0
5        min_cumulative_sum = float('inf')
6      
7        # Iterate through the array to find the minimum cumulative sum
8        for num in nums:
9            cumulative_sum += num
10            min_cumulative_sum = min(min_cumulative_sum, cumulative_sum)
11      
12        # Calculate the minimum start value
13        # If min_cumulative_sum is negative, we need 1 - min_cumulative_sum to ensure step sum >= 1
14        # If min_cumulative_sum is non-negative, we just need start value = 1
15        return max(1, 1 - min_cumulative_sum)
16
1class Solution {
2    public int minStartValue(int[] nums) {
3        // Track the running sum
4        int runningSum = 0;
5      
6        // Track the minimum prefix sum encountered
7        int minPrefixSum = Integer.MAX_VALUE;
8      
9        // Iterate through the array and calculate prefix sums
10        for (int num : nums) {
11            // Update running sum
12            runningSum += num;
13          
14            // Update minimum prefix sum if current sum is smaller
15            minPrefixSum = Math.min(minPrefixSum, runningSum);
16        }
17      
18        // Calculate minimum start value
19        // If minPrefixSum is negative or zero, we need (1 - minPrefixSum) to ensure sum >= 1
20        // If minPrefixSum is positive, start value of 1 is sufficient
21        return Math.max(1, 1 - minPrefixSum);
22    }
23}
24
1class Solution {
2public:
3    int minStartValue(vector<int>& nums) {
4        // Track the running sum of the array
5        int runningSum = 0;
6      
7        // Track the minimum value reached by the running sum
8        int minSum = INT_MAX;
9      
10        // Iterate through each number and calculate prefix sums
11        for (int num : nums) {
12            runningSum += num;
13            minSum = min(minSum, runningSum);
14        }
15      
16        // The start value must ensure that (startValue + minSum) >= 1
17        // This means startValue >= 1 - minSum
18        // Also, startValue must be at least 1
19        return max(1, 1 - minSum);
20    }
21};
22
1/**
2 * Finds the minimum positive starting value such that the cumulative sum
3 * never drops below 1 at any point during iteration through the array.
4 * 
5 * @param nums - Array of integers (positive or negative)
6 * @returns The minimum positive starting value
7 */
8function minStartValue(nums: number[]): number {
9    // Track the running cumulative sum
10    let cumulativeSum: number = 0;
11  
12    // Track the minimum cumulative sum encountered
13    let minCumulativeSum: number = Infinity;
14  
15    // Iterate through each number in the array
16    for (const currentNumber of nums) {
17        // Update the cumulative sum
18        cumulativeSum += currentNumber;
19      
20        // Update the minimum cumulative sum if current is lower
21        minCumulativeSum = Math.min(minCumulativeSum, cumulativeSum);
22    }
23  
24    // Calculate the minimum starting value
25    // If minCumulativeSum is negative or zero, we need (1 - minCumulativeSum) to ensure sum >= 1
26    // If minCumulativeSum is positive, starting value of 1 is sufficient
27    return Math.max(1, 1 - minCumulativeSum);
28}
29

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input array nums. The algorithm iterates through the array exactly once, performing constant-time operations (addition, comparison, and assignment) for each element.

Space Complexity: O(1). The algorithm uses only a fixed amount of extra space regardless of the input size. It maintains two variables s (running sum) and t (minimum prefix sum), which require constant space. The input array is not modified and no additional data structures are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Initialize the Minimum to Infinity

A common mistake is initializing min_cumulative_sum to 0 instead of infinity. This causes incorrect results when all cumulative sums are positive.

Incorrect:

min_cumulative_sum = 0  # Wrong initialization

Why it fails: If nums = [1, 2, 3], all cumulative sums are positive (1, 3, 6). With min_cumulative_sum = 0, the result would be max(1, 1 - 0) = 1, which is correct by coincidence. However, the logic is flawed because 0 was never actually a cumulative sum value.

Correct:

min_cumulative_sum = float('inf')  # Correct initialization

2. Confusing Minimum Sum with Minimum Element

Some might track the minimum array element instead of the minimum cumulative sum.

Incorrect:

min_element = min(nums)
return max(1, 1 - min_element)  # Wrong approach

Why it fails: For nums = [10, -20, 30], the minimum element is -20, giving max(1, 21) = 21. But the cumulative sums are [10, -10, 20], with minimum -10, so the correct answer is max(1, 11) = 11.

3. Off-by-One Error in Final Calculation

Incorrectly calculating the starting value by forgetting the "+1" adjustment.

Incorrect:

return max(1, -min_cumulative_sum)  # Missing the +1

Why it fails: If min_cumulative_sum = -4, this returns max(1, 4) = 4. But starting with 4 gives a minimum sum of 0, which violates the "at least 1" requirement. The correct formula is 1 - min_cumulative_sum.

4. Not Handling the Case When All Sums Are Positive

Forgetting the max(1, ...) wrapper and returning a non-positive starting value.

Incorrect:

return 1 - min_cumulative_sum  # Missing max(1, ...)

Why it fails: If nums = [5, 10], min_cumulative_sum = 5, leading to 1 - 5 = -4. A negative starting value violates the problem constraints. The max(1, ...) ensures the result is always positive.

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Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
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1public static int binarySearch(int[] arr, int target) {
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4
5    while (left ___ right) {
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16
1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
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7        if (arr[mid] == target) return mid;
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9            ___ = mid + 1;
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11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

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