LeetCode #2197 — HARD

Replace Non-Coprime Numbers in Array

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 integers nums. Perform the following steps:

Find any two adjacent numbers in nums that are non-coprime.
If no such numbers are found, stop the process.
Otherwise, delete the two numbers and replace them with their LCM (Least Common Multiple).
Repeat this process as long as you keep finding two adjacent non-coprime numbers.

Return the final modified array. It can be shown that replacing adjacent non-coprime numbers in any arbitrary order will lead to the same result.

The test cases are generated such that the values in the final array are less than or equal to 10⁸.

Two values x and y are non-coprime if GCD(x, y) > 1 where GCD(x, y) is the Greatest Common Divisor of x and y.

Example 1:

Input: nums = [6,4,3,2,7,6,2]
Output: [12,7,6]
Explanation: 
- (6, 4) are non-coprime with LCM(6, 4) = 12. Now, nums = [12,3,2,7,6,2].
- (12, 3) are non-coprime with LCM(12, 3) = 12. Now, nums = [12,2,7,6,2].
- (12, 2) are non-coprime with LCM(12, 2) = 12. Now, nums = [12,7,6,2].
- (6, 2) are non-coprime with LCM(6, 2) = 6. Now, nums = [12,7,6].
There are no more adjacent non-coprime numbers in nums.
Thus, the final modified array is [12,7,6].
Note that there are other ways to obtain the same resultant array.

Example 2:

Input: nums = [2,2,1,1,3,3,3]
Output: [2,1,1,3]
Explanation: 
- (3, 3) are non-coprime with LCM(3, 3) = 3. Now, nums = [2,2,1,1,3,3].
- (3, 3) are non-coprime with LCM(3, 3) = 3. Now, nums = [2,2,1,1,3].
- (2, 2) are non-coprime with LCM(2, 2) = 2. Now, nums = [2,1,1,3].
There are no more adjacent non-coprime numbers in nums.
Thus, the final modified array is [2,1,1,3].
Note that there are other ways to obtain the same resultant array.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
The test cases are generated such that the values in the final array are less than or equal to 10⁸.

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums. Perform the following steps: Find any two adjacent numbers in nums that are non-coprime. If no such numbers are found, stop the process. Otherwise, delete the two numbers and replace them with their LCM (Least Common Multiple). Repeat this process as long as you keep finding two adjacent non-coprime numbers. Return the final modified array. It can be shown that replacing adjacent non-coprime numbers in any arbitrary order will lead to the same result. The test cases are generated such that the values in the final array are less than or equal to 108. Two values x and y are non-coprime if GCD(x, y) > 1 where GCD(x, y) is the Greatest Common Divisor of x and y.

Baseline thinking

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

Pattern signal: Array · Math · Stack

Example 1

[6,4,3,2,7,6,2]

Example 2

[2,2,1,1,3,3,3]

Core Insight

What unlocks the optimal approach

Notice that the order of merging two numbers into their LCM does not matter so we can greedily merge elements to its left if possible.
If a new value is formed, we should recursively check if it can be merged with the value to its left.
To simulate the merge efficiently, we can maintain a stack that stores processed elements. When we iterate through the array, we only compare with the top of the stack (which is the value to its left).

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 #2197: Replace Non-Coprime Numbers in Array
class Solution {
    public List<Integer> replaceNonCoprimes(int[] nums) {
        List<Integer> stk = new ArrayList<>();
        for (int x : nums) {
            stk.add(x);
            while (stk.size() > 1) {
                x = stk.get(stk.size() - 1);
                int y = stk.get(stk.size() - 2);
                int g = gcd(x, y);
                if (g == 1) {
                    break;
                }
                stk.remove(stk.size() - 1);
                stk.set(stk.size() - 1, (int) ((long) x * y / g));
            }
        }
        return stk;
    }

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

// Accepted solution for LeetCode #2197: Replace Non-Coprime Numbers in Array
func replaceNonCoprimes(nums []int) []int {
	stk := []int{}
	for _, x := range nums {
		stk = append(stk, x)
		for len(stk) > 1 {
			x = stk[len(stk)-1]
			y := stk[len(stk)-2]
			g := gcd(x, y)
			if g == 1 {
				break
			}
			stk = stk[:len(stk)-1]
			stk[len(stk)-1] = x * y / g
		}
	}
	return stk
}

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

# Accepted solution for LeetCode #2197: Replace Non-Coprime Numbers in Array
class Solution:
    def replaceNonCoprimes(self, nums: List[int]) -> List[int]:
        stk = []
        for x in nums:
            stk.append(x)
            while len(stk) > 1:
                x, y = stk[-2:]
                g = gcd(x, y)
                if g == 1:
                    break
                stk.pop()
                stk[-1] = x * y // g
        return stk

// Accepted solution for LeetCode #2197: Replace Non-Coprime Numbers in Array
impl Solution {
    pub fn replace_non_coprimes(nums: Vec<i32>) -> Vec<i32> {
        fn gcd(mut a: i64, mut b: i64) -> i64 {
            while b != 0 {
                let t = a % b;
                a = b;
                b = t;
            }
            a
        }

        let mut stk: Vec<i64> = Vec::new();
        for x in nums {
            stk.push(x as i64);
            while stk.len() > 1 {
                let x = *stk.last().unwrap();
                let y = stk[stk.len() - 2];
                let g = gcd(x, y);
                if g == 1 {
                    break;
                }
                stk.pop();
                let last = stk.last_mut().unwrap();
                *last = x / g * y;
            }
        }

        stk.into_iter().map(|v| v as i32).collect()
    }
}

// Accepted solution for LeetCode #2197: Replace Non-Coprime Numbers in Array
function replaceNonCoprimes(nums: number[]): number[] {
    const gcd = (a: number, b: number): number => {
        if (b === 0) {
            return a;
        }
        return gcd(b, a % b);
    };
    const stk: number[] = [];
    for (let x of nums) {
        stk.push(x);
        while (stk.length > 1) {
            x = stk.at(-1)!;
            const y = stk.at(-2)!;
            const g = gcd(x, y);
            if (g === 1) {
                break;
            }
            stk.pop();
            stk.pop();
            stk.push(((x * y) / g) | 0);
        }
    }
    return stk;
}

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 M)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.