LeetCode #3574 — HARD

Maximize Subarray GCD Score

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array of positive integers nums and an integer k.

You may perform at most k operations. In each operation, you can choose one element in the array and double its value. Each element can be doubled at most once.

The score of a contiguous subarray is defined as the product of its length and the greatest common divisor (GCD) of all its elements.

Your task is to return the maximum score that can be achieved by selecting a contiguous subarray from the modified array.

Note:

The greatest common divisor (GCD) of an array is the largest integer that evenly divides all the array elements.

Example 1:

Input: nums = [2,4], k = 1

Output: 8

Explanation:

Double nums[0] to 4 using one operation. The modified array becomes [4, 4].
The GCD of the subarray [4, 4] is 4, and the length is 2.
Thus, the maximum possible score is 2 × 4 = 8.

Example 2:

Input: nums = [3,5,7], k = 2

Output: 14

Explanation:

Double nums[2] to 14 using one operation. The modified array becomes [3, 5, 14].
The GCD of the subarray [14] is 14, and the length is 1.
Thus, the maximum possible score is 1 × 14 = 14.

Example 3:

Input: nums = [5,5,5], k = 1

Output: 15

Explanation:

The subarray [5, 5, 5] has a GCD of 5, and its length is 3.
Since doubling any element doesn't improve the score, the maximum score is 3 × 5 = 15.

Constraints:

1 <= n == nums.length <= 1500
1 <= nums[i] <= 10⁹
1 <= k <= n

Step 01

Brute Force Baseline

Problem summary: You are given an array of positive integers nums and an integer k. You may perform at most k operations. In each operation, you can choose one element in the array and double its value. Each element can be doubled at most once. The score of a contiguous subarray is defined as the product of its length and the greatest common divisor (GCD) of all its elements. Your task is to return the maximum score that can be achieved by selecting a contiguous subarray from the modified array. Note: The greatest common divisor (GCD) of an array is the largest integer that evenly divides all the array elements.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[2,4]
1

Example 2

[3,5,7]
2

Example 3

[5,5,5]
1

Step 02

Core Insight

What unlocks the optimal approach

Try iterating over the subarrays
Handle the 2s in the factors of elements separately

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

Largest constraint values

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 #3574: Maximize Subarray GCD Score
class Solution {
    public long maxGCDScore(int[] nums, int k) {
        int n = nums.length;
        int[] cnt = new int[n];
        for (int i = 0; i < n; ++i) {
            for (int x = nums[i]; x % 2 == 0; x /= 2) {
                ++cnt[i];
            }
        }
        long ans = 0;
        for (int l = 0; l < n; ++l) {
            int g = 0;
            int mi = 1 << 30;
            int t = 0;
            for (int r = l; r < n; ++r) {
                g = gcd(g, nums[r]);
                if (cnt[r] < mi) {
                    mi = cnt[r];
                    t = 1;
                } else if (cnt[r] == mi) {
                    ++t;
                }
                ans = Math.max(ans, (r - l + 1L) * (t > k ? g : g * 2));
            }
        }
        return ans;
    }

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

// Accepted solution for LeetCode #3574: Maximize Subarray GCD Score
func maxGCDScore(nums []int, k int) int64 {
	n := len(nums)
	cnt := make([]int, n)
	for i, x := range nums {
		for x%2 == 0 {
			cnt[i]++
			x /= 2
		}
	}

	ans := 0
	for l := 0; l < n; l++ {
		g := 0
		mi := math.MaxInt32
		t := 0
		for r := l; r < n; r++ {
			g = gcd(g, nums[r])
			if cnt[r] < mi {
				mi = cnt[r]
				t = 1
			} else if cnt[r] == mi {
				t++
			}
			length := r - l + 1
			score := g * length
			if t <= k {
				score *= 2
			}
			ans = max(ans, score)
		}
	}

	return int64(ans)
}

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

# Accepted solution for LeetCode #3574: Maximize Subarray GCD Score
class Solution:
    def maxGCDScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        cnt = [0] * n
        for i, x in enumerate(nums):
            while x % 2 == 0:
                cnt[i] += 1
                x //= 2
        ans = 0
        for l in range(n):
            g = 0
            mi = inf
            t = 0
            for r in range(l, n):
                g = gcd(g, nums[r])
                if cnt[r] < mi:
                    mi = cnt[r]
                    t = 1
                elif cnt[r] == mi:
                    t += 1
                ans = max(ans, (g if t > k else g * 2) * (r - l + 1))
        return ans

// Accepted solution for LeetCode #3574: Maximize Subarray GCD Score
impl Solution {
    pub fn max_gcd_score(nums: Vec<i32>, k: i32) -> i64 {
        let n = nums.len();
        let mut cnt = vec![0i32; n];
        for i in 0..n {
            let mut x = nums[i];
            while x % 2 == 0 {
                cnt[i] += 1;
                x /= 2;
            }
        }

        let mut ans: i64 = 0;
        for l in 0..n {
            let mut g: i32 = 0;
            let mut mi: i32 = 1 << 30;
            let mut t: i32 = 0;
            for r in l..n {
                g = Self::gcd(g, nums[r]);
                if cnt[r] < mi {
                    mi = cnt[r];
                    t = 1;
                } else if cnt[r] == mi {
                    t += 1;
                }
                let val = if t > k { g as i64 } else { (g * 2) as i64 };
                ans = ans.max((r as i64 - l as i64 + 1) * val);
            }
        }
        ans
    }

    fn gcd(a: i32, b: i32) -> i32 {
        if b == 0 { a } else { Self::gcd(b, a % b) }
    }
}

// Accepted solution for LeetCode #3574: Maximize Subarray GCD Score
function maxGCDScore(nums: number[], k: number): number {
    const n = nums.length;
    const cnt: number[] = Array(n).fill(0);

    for (let i = 0; i < n; ++i) {
        let x = nums[i];
        while (x % 2 === 0) {
            cnt[i]++;
            x /= 2;
        }
    }

    let ans = 0;
    for (let l = 0; l < n; ++l) {
        let g = 0;
        let mi = Number.MAX_SAFE_INTEGER;
        let t = 0;
        for (let r = l; r < n; ++r) {
            g = gcd(g, nums[r]);
            if (cnt[r] < mi) {
                mi = cnt[r];
                t = 1;
            } else if (cnt[r] === mi) {
                t++;
            }
            const len = r - l + 1;
            const score = (t > k ? g : g * 2) * len;
            ans = Math.max(ans, score);
        }
    }

    return ans;
}

function gcd(a: number, b: number): number {
    while (b !== 0) {
        const temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

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^2 × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.