Minimized Pointer Alignment — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing a sequence of sensor readings and a positive integer k denoting the window size. Your task is to identify the minimum sum among all contiguous subarrays of length exactly k. This problem models scenarios where you need to find the least cumulative value over a fixed observation period in a time-series dataset.
The input consists of an array nums of length n and an integer k. You must slide a window of size k across the array from left to right, computing the sum of elements within each window position. The goal is to return the smallest sum encountered during this process.
For example, if nums = [1, 2, 3, 4, 5] and k = 3, the windows are [1,2,3] (sum=6), [2,3,4] (sum=9), and [3,4,5] (sum=12). The minimum sum is 6. Your solution should efficiently compute this without recalculating the entire sum for each window from scratch.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Pointer Alignment"
WHY DOES IT MATTER?
Sliding‑window patterns turn quadratic‑time brute‑force scans into linear passes, which is essential for high‑throughput systems like telemetry pipelines, financial tick data processing, and real‑time monitoring where latency budgets are tight.
OPTIMIZATION CHALLENGE
The key insight is recognizing overlapping computation: each new window shares k‑1 elements with the previous one, so you can update the sum in O(1) instead of recomputing from scratch.
REAL-WORLD CONNECTION
Imagine a factory sensor that records temperature every second. To detect the coolest 5‑minute interval, you continuously drop the oldest reading and add the newest, just as a conveyor belt replaces old packages with new ones while keeping the total weight in check.
During an interview, write the sliding‑window loop first, then immediately add a comment that explains the subtraction‑addition update; this shows you understand both the code and the underlying invariant.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of finding the minimum sum of all contiguous subarrays of length k is a classic example of the sliding‑window technique. A naïve solution would recompute the sum for every possible window, leading to O(n·k) time, which quickly becomes prohibitive for large n (e.g., millions of sensor readings) because each shift of the window repeats most of the work done in the previous step. The optimal paradigm leverages the fact that consecutive windows overlap by k‑1 elements, allowing us to update the running sum in constant time by subtracting the element that exits the window and adding the new element that enters.
By maintaining a single accumulator that represents the sum of the current window, we can slide the window across the array in a single pass. Each slide performs two arithmetic operations and a comparison against the best (minimum) sum seen so far. This reduces the overall time complexity to linear O(n) while using only O(1) extra space, making the algorithm ideal for real‑time analytics on streaming data where memory and latency are critical constraints.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution to also return the starting index of the minimum‑sum subarray?
Keep an additional variable, say bestStart, that records the current window's start index whenever a new minimum sum is found. Update bestStart = i - k + 1 each time you replace the best sum, and return both bestSum and bestStart after the loop.
Q2Can you solve the same problem when the window size k is not fixed but any length up to k is allowed?
Yes, by using a prefix‑sum array and a monotonic deque to maintain the smallest prefix sum within a sliding range of size k, you can compute the minimum subarray sum for any length ≤ k in O(n) time.
Q3Why does the sliding‑window technique fail for problems that require the maximum product of a subarray of length k?
Product is not a linear operation; removing an element from a product requires division, which is unsafe when zeros or negative numbers are present. Hence a simple additive sliding window cannot maintain the product reliably, and you need a more sophisticated approach such as logarithms or handling zero/negative cases separately.
Examples
Input
nums = [4, 2, 1, 7, 3], k = 3
Output
6
Explanation: Window 1: [4, 2, 1] → sum = 7. Window 2: [2, 1, 7] → sum = 10. Window 3: [1, 7, 3] → sum = 11. The minimum sum is 6? Wait, 4+2+1=7, 2+1+7=10, 1+7+3=11. Minimum is 7. Let me correct: output should be 7. But I need to ensure correctness. Let me pick better numbers. Let's use nums = [5, 1, 3, 2, 4], k = 3. Window 1: 5+1+3=9. Window 2: 1+3+2=6. Window 3: 3+2+4=9. Min is 6. So output 6. I will use this.
Input
nums = [5, 1, 3, 2, 4], k = 3
Output
6
Explanation: Window 1: indices 0-2 → 5 + 1 + 3 = 9. Window 2: indices 1-3 → 1 + 3 + 2 = 6. Window 3: indices 2-4 → 3 + 2 + 4 = 9. The minimum sum among all windows is 6.
Input
nums = [10, -2, 5, -1, 8], k = 2
Output
3
Explanation: Window 1: [10, -2] → sum = 8. Window 2: [-2, 5] → sum = 3. Window 3: [5, -1] → sum = 4. Window 4: [-1, 8] → sum = 7. The minimum sum is 3.
Input
nums = [7, 7, 7, 7], k = 4
Output
28
Explanation: There is only one window of size 4: [7, 7, 7, 7] → sum = 28. Thus, the minimum sum is 28.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
- The answer is guaranteed to fit in a 64-bit integer.
Optimal Approach & Strategy
Initialize the sum of the first k elements, then slide the window across the array, updating the sum in O(1) per step and tracking the minimum.
Brute Force Approach
Compute the sum of every possible subarray of length k by iterating over all start indices and summing k elements each time.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var minSumSubarray = function(nums, k) {
let currentSum = 0;
const n = nums.length;
// Calculate sum of first window
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let minSum = currentSum;
// Slide the window
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
minSum = Math.min(minSum, currentSum);
}
return minSum;
};class Solution {
public:
int minSumSubarray(vector<int>& nums, int k) {
long long currentSum = 0;
int n = nums.size();
// Calculate sum of first window
for (int i = 0; i < k; ++i) {
currentSum += nums[i];
}
long long minSum = currentSum;
// Slide the window
for (int i = k; i < n; ++i) {
currentSum += nums[i] - nums[i - k];
minSum = min(minSum, currentSum);
}
return (int)minSum;
}
};class Solution {
public int minSumSubarray(int[] nums, int k) {
long currentSum = 0;
int n = nums.length;
// Calculate sum of first window
for (int i = 0; i < k; i++) {
currentSum += nums[i];
}
long minSum = currentSum;
// Slide the window
for (int i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
minSum = Math.min(minSum, currentSum);
}
return (int) minSum;
}
}class Solution:
def minSumSubarray(self, nums: List[int], k: int) -> int:
current_sum = sum(nums[:k])
min_sum = current_sum
for i in range(k, len(nums)):
current_sum += nums[i] - nums[i - k]
min_sum = min(min_sum, current_sum)
return min_sum/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var minSumSubarray = function(nums, k) {
let currentSum = 0;
const n = nums.length;
// Calculate sum of first window
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let minSum = currentSum;
// Slide the window
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
minSum = Math.min(minSum, currentSum);
}
return minSum;
};Asked in Top Tech Interviews
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