Minimized Interval Partition — Problem Statement & Solution Guide
Problem Description
Given an array of integers nums and a positive integer k, determine the sum of every contiguous subarray of length k. The result should be an array where the element at index i represents the sum of the subarray starting at i and ending at i + k - 1. This computation must be performed efficiently by leveraging the relationship between consecutive window sums, specifically by sliding the window one position at a time and updating the total in constant time rather than recalculating from scratch for each position.
The core logic involves initializing the sum of the first k elements, then for each subsequent window, subtracting the element that exits the window (the leftmost element of the previous window) and adding the new element that enters the window (the rightmost element of the current window). This approach ensures that each element is processed a constant number of times, leading to an optimal linear time complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Interval Partition"
WHY DOES IT MATTER?
Sliding window is a fundamental pattern for problems involving contiguous sequences where adjacent windows share most of their elements. Recognizing this overlap lets you transform quadratic‑time brute‑force solutions into linear‑time ones, a skill that differentiates senior candidates.
OPTIMIZATION CHALLENGE
The key insight is the recurrence relation: windowSum[i+1] = windowSum[i] - nums[i] + nums[i+k]. This reduces the per‑window work from O(k) to O(1), collapsing the overall complexity from O(n·k) to O(n).
REAL-WORLD CONNECTION
Think of a moving sensor that records temperature every second; to compute the average temperature over the last k seconds continuously, you don't recalculate the whole sum each second—you just drop the oldest reading and add the newest, exactly like a sliding window.
During an interview, write the initial O(k) sum for the first window explicitly, then immediately transition to the loop that updates the sum in constant time. This shows you understand both the baseline and the optimization.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The sliding‑window technique exploits the overlap between consecutive subarrays of fixed length. For a window of size k, the sum of the window starting at index i+1 can be derived from the sum at i by subtracting nums[i] (the element exiting the window) and adding nums[i+k] (the new element entering). This recurrence eliminates the need to recompute the sum from scratch for each position. A naive solution would iterate over every possible start index and sum k elements each time, leading to O(n·k) time, which becomes prohibitive when n and k are large (e.g., n = 10^6, k = 10^5). By maintaining a running total and updating it in O(1) per step, the overall algorithm runs in linear time O(n), which is optimal because every element must be inspected at least once. The space requirement is also minimal: only the result array of size n‑k+1 and a few scalar variables are needed, yielding O(n) total auxiliary space.
Interview Questions on This Problem
Q1How would you compute the sum of every subarray of length k in an array of size n in O(n) time?
Initialize the sum of the first k elements, store it, then slide the window: for each i from 1 to n‑k, update sum = sum - nums[i‑1] + nums[i+k‑1] and store the new sum. This runs in O(n) time and O(1) extra space besides the output.
Q2What modifications are needed to return the maximum sum of any subarray of length k instead of all sums?
Apply the same sliding‑window update, but keep a variable maxSum that tracks the largest window sum seen so far. After each update, compare and possibly replace maxSum. The algorithm remains O(n) time and O(1) extra space.
Q3Can the sliding‑window approach be extended to compute the average of each subarray of length k? Explain any pitfalls.
Yes. Compute the window sum as before, then divide by k to get the average for each position. The pitfall is integer division in languages like Java or C++; you must cast to a floating‑point type before division to avoid truncation.
Examples
Input
nums = [1, 2, 3, 4, 5], k = 3
Output
[6, 9, 12]
Explanation: 1. Initialize sum of first window [1, 2, 3] as 6. Result: [6]. 2. Slide window to [2, 3, 4]: subtract 1, add 4 -> 6 - 1 + 4 = 9. Result: [6, 9]. 3. Slide window to [3, 4, 5]: subtract 2, add 5 -> 9 - 2 + 5 = 12. Result: [6, 9, 12].
Input
nums = [10, -5, 20, 15, 5], k = 2
Output
[5, 15, 35, 20]
Explanation: 1. First window [10, -5] sum is 5. Result: [5]. 2. Next window [-5, 20]: subtract 10, add 20 -> 5 - 10 + 20 = 15. Result: [5, 15]. 3. Next window [20, 15]: subtract -5, add 15 -> 15 - (-5) + 15 = 35. Result: [5, 15, 35]. 4. Next window [15, 5]: subtract 20, add 5 -> 35 - 20 + 5 = 20. Result: [5, 15, 35, 20].
Input
nums = [7, 7, 7, 7], k = 4
Output
[28]
Explanation: 1. The array length is 4 and k is 4, so there is only one possible window covering the entire array. 2. Sum of [7, 7, 7, 7] is 28. Result: [28].
Input
nums = [1, 1, 1, 1, 1, 1], k = 1
Output
[1, 1, 1, 1, 1, 1]
Explanation: 1. Since k=1, each window consists of a single element. 2. The sum of each single-element window is the element itself. 3. Result: [1, 1, 1, 1, 1, 1].
Constraints
- 1 <= nums.length <= 10^5
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Compute the first window sum once, then slide the window across the array, updating the sum by removing the leftmost element and adding the new rightmost element in O(1) per step, achieving O(n) time.
Brute Force Approach
For each possible start index, sum the next k elements by iterating over them, resulting in O(n·k) time.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var slidingWindowSum = function(nums, k) {
const n = nums.length;
if (n < k) return [];
let currentSum = 0;
const result = [];
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
result.push(currentSum);
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
result.push(currentSum);
}
return result;
};class Solution {
public:
vector<int> slidingWindowSum(vector<int>& nums, int k) {
int n = nums.size();
vector<int> result;
if (n < k) return result;
long long currentSum = 0;
for (int i = 0; i < k; ++i) {
currentSum += nums[i];
}
result.push_back(currentSum);
for (int i = k; i < n; ++i) {
currentSum += nums[i] - nums[i - k];
result.push_back(currentSum);
}
return result;
}
};class Solution {
public int[] slidingWindowSum(int[] nums, int k) {
int n = nums.length;
if (n < k) return new int[0];
long currentSum = 0;
int[] result = new int[n - k + 1];
for (int i = 0; i < k; i++) {
currentSum += nums[i];
}
result[0] = (int) currentSum;
for (int i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
result[i - k + 1] = (int) currentSum;
}
return result;
}
}class Solution:
def slidingWindowSum(self, nums: List[int], k: int) -> List[int]:
n = len(nums)
if n < k:
return []
current_sum = sum(nums[:k])
result = [current_sum]
for i in range(k, n):
current_sum += nums[i] - nums[i - k]
result.append(current_sum)
return result/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var slidingWindowSum = function(nums, k) {
const n = nums.length;
if (n < k) return [];
let currentSum = 0;
const result = [];
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
result.push(currentSum);
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
result.push(currentSum);
}
return result;
};Asked in Top Tech Interviews
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