LeetCode #2607 — MEDIUM

Make K-Subarray Sums Equal

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed integer array arr and an integer k. The array arr is circular. In other words, the first element of the array is the next element of the last element, and the last element of the array is the previous element of the first element.

You can do the following operation any number of times:

Pick any element from arr and increase or decrease it by 1.

Return the minimum number of operations such that the sum of each subarray of length k is equal.

A subarray is a contiguous part of the array.

Example 1:

Input: arr = [1,4,1,3], k = 2
Output: 1
Explanation: we can do one operation on index 1 to make its value equal to 3.
The array after the operation is [1,3,1,3]
- Subarray starts at index 0 is [1, 3], and its sum is 4 
- Subarray starts at index 1 is [3, 1], and its sum is 4 
- Subarray starts at index 2 is [1, 3], and its sum is 4 
- Subarray starts at index 3 is [3, 1], and its sum is 4

Example 2:

Input: arr = [2,5,5,7], k = 3
Output: 5
Explanation: we can do three operations on index 0 to make its value equal to 5 and two operations on index 3 to make its value equal to 5.
The array after the operations is [5,5,5,5]
- Subarray starts at index 0 is [5, 5, 5], and its sum is 15
- Subarray starts at index 1 is [5, 5, 5], and its sum is 15
- Subarray starts at index 2 is [5, 5, 5], and its sum is 15
- Subarray starts at index 3 is [5, 5, 5], and its sum is 15

Constraints:

1 <= k <= arr.length <= 10⁵
1 <= arr[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array arr and an integer k. The array arr is circular. In other words, the first element of the array is the next element of the last element, and the last element of the array is the previous element of the first element. You can do the following operation any number of times: Pick any element from arr and increase or decrease it by 1. Return the minimum number of operations such that the sum of each subarray of length k is equal. A subarray is a contiguous part of the array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Greedy

Example 1

[1,4,1,3]
2

Example 2

[2,5,5,7]
3

Core Insight

What unlocks the optimal approach

Think about gcd(n, k). How will it help to calculate the answer?
indices i and j are in the same group if gcd(n, k) mod i = gcd(n, k) mod j. Each group should have equal elements. Think about the minimum number of operations for each group
The minimum number of operations for each group equals the summation of differences between the elements and the median of elements inside the group.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
class Solution {
    public long makeSubKSumEqual(int[] arr, int k) {
        int n = arr.length;
        int g = gcd(n, k);
        long ans = 0;
        for (int i = 0; i < g; ++i) {
            List<Integer> t = new ArrayList<>();
            for (int j = i; j < n; j += g) {
                t.add(arr[j]);
            }
            t.sort((a, b) -> a - b);
            int mid = t.get(t.size() >> 1);
            for (int x : t) {
                ans += Math.abs(x - mid);
            }
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
func makeSubKSumEqual(arr []int, k int) (ans int64) {
	n := len(arr)
	g := gcd(n, k)
	for i := 0; i < g; i++ {
		t := []int{}
		for j := i; j < n; j += g {
			t = append(t, arr[j])
		}
		sort.Ints(t)
		mid := t[len(t)/2]
		for _, x := range t {
			ans += int64(abs(x - mid))
		}
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
class Solution:
    def makeSubKSumEqual(self, arr: List[int], k: int) -> int:
        n = len(arr)
        g = gcd(n, k)
        ans = 0
        for i in range(g):
            t = sorted(arr[i:n:g])
            mid = t[len(t) >> 1]
            ans += sum(abs(x - mid) for x in t)
        return ans

// Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
// class Solution {
//     public long makeSubKSumEqual(int[] arr, int k) {
//         int n = arr.length;
//         int g = gcd(n, k);
//         long ans = 0;
//         for (int i = 0; i < g; ++i) {
//             List<Integer> t = new ArrayList<>();
//             for (int j = i; j < n; j += g) {
//                 t.add(arr[j]);
//             }
//             t.sort((a, b) -> a - b);
//             int mid = t.get(t.size() >> 1);
//             for (int x : t) {
//                 ans += Math.abs(x - mid);
//             }
//         }
//         return ans;
//     }
// 
//     private int gcd(int a, int b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

// Accepted solution for LeetCode #2607: Make K-Subarray Sums Equal
function makeSubKSumEqual(arr: number[], k: number): number {
    const n = arr.length;
    const g = gcd(n, k);
    let ans = 0;
    for (let i = 0; i < g; ++i) {
        const t: number[] = [];
        for (let j = i; j < n; j += g) {
            t.push(arr[j]);
        }
        t.sort((a, b) => a - b);
        const mid = t[t.length >> 1];
        for (const x of t) {
            ans += Math.abs(x - mid);
        }
    }
    return ans;
}

function gcd(a: number, b: number): number {
    if (b === 0) {
        return a;
    }
    return gcd(b, a % b);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.