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Hard
2057
Weekly Contest 283 Q4
Stack
Array
Math
Number Theory

2197. Replace Non-Coprime Numbers in Array

中文文档

Description

You are given an array of integers nums. Perform the following steps:

  1. Find any two adjacent numbers in nums that are non-coprime.
  2. If no such numbers are found, stop the process.
  3. Otherwise, delete the two numbers and replace them with their LCM (Least Common Multiple).
  4. 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.

Solutions

Solution 1: Stack

If there exist three adjacent numbers $x$, $y$, $z$ that can be merged, then the result of first merging $x$ and $y$, then merging $z$, is the same as the result of first merging $y$ and $z$, then merging $x$. Both results are $\textit{LCM}(x, y, z)$.

Therefore, we can always prefer to merge the adjacent numbers on the left, and then merge the result with the adjacent number on the right.

We use a stack to simulate this process. We traverse the array, and for each number, we push it into the stack. Then we continuously check whether the top two numbers of the stack are coprime. If they are not coprime, we pop these two numbers out of the stack, and then push their least common multiple into the stack, until the top two numbers of the stack are coprime, or there are less than two elements in the stack.

The final elements in the stack are the final result.

The time complexity is $O(n \times \log M)$, and the space complexity is $O(n)$. Where $M$ is the maximum value in the array.

Python3

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

Java

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);
    }
}

C++

class Solution {
public:
    vector<int> replaceNonCoprimes(vector<int>& nums) {
        vector<int> stk;
        for (int x : nums) {
            stk.push_back(x);
            while (stk.size() > 1) {
                x = stk.back();
                int y = stk[stk.size() - 2];
                int g = __gcd(x, y);
                if (g == 1) {
                    break;
                }
                stk.pop_back();
                stk.back() = 1LL * x * y / g;
            }
        }
        return stk;
    }
};

Go

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)
}

TypeScript

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;
}