Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1769.html

1769. Minimum Number of Operations to Move All Balls to Each Box

Level

Medium

Description

You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the i-th box is empty, and '1' if it contains one ball.

In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes.

Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the i-th box.

Each answer[i] is calculated considering the initial state of the boxes.

Example 1:

Input: boxes = “110”

Output: [1,1,3]

Explanation: The answer for each box is as follows:

1) First box: you will have to move one ball from the second box to the first box in one operation.

2) Second box: you will have to move one ball from the first box to the second box in one operation.

3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball from the second box to the third box in one operation.

Example 2:

Input: boxes = “001011”

Output: [11,8,5,4,3,4]

Constraints:

  • n == boxes.length
  • 1 <= n <= 2000
  • boxes[i] is either '0' or '1'.

Solution

For each index of boxes, calculate the number of balls to the left of the index and the number of balls to the right of the index. Then for each index, calculate the number of operations to move all balls to the left to the current index, and the number of operations to move all balls to the right to the current index.

  • class Solution {
    public int[] minOperations(String boxes) {
        int length = boxes.length();
        int[] leftBalls = new int[length];
        int[] rightBalls = new int[length];
        for (int i = 1; i < length; i++)
            leftBalls[i] = leftBalls[i - 1] + boxes.charAt(i - 1) - '0';
        for (int i = length - 2; i >= 0; i--)
            rightBalls[i] = rightBalls[i + 1] + boxes.charAt(i + 1) - '0';
        int[] leftOperations = new int[length];
        int[] rightOperations = new int[length];
        for (int i = 1; i < length; i++)
            leftOperations[i] = leftOperations[i - 1] + leftBalls[i];
        for (int i = length - 2; i >= 0; i--)
            rightOperations[i] = rightOperations[i + 1] + rightBalls[i];
        int[] operations = new int[length];
        for (int i = 0; i < length; i++)
            operations[i] = leftOperations[i] + rightOperations[i];
        return operations;
    }
}

############

class Solution {
    public int[] minOperations(String boxes) {
        int n = boxes.length();
        int[] ans = new int[n];
        for (int i = 1, cnt = 0; i < n; ++i) {
            if (boxes.charAt(i - 1) == '1') {
                ++cnt;
            }
            ans[i] = ans[i - 1] + cnt;
        }
        for (int i = n - 2, cnt = 0, s = 0; i >= 0; --i) {
            if (boxes.charAt(i + 1) == '1') {
                ++cnt;
            }
            s += cnt;
            ans[i] += s;
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    vector<int> minOperations(string s) {
        int N = s.size();
        vector<int> ans(s.size());
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                if (s[j] == '1') ans[i] += abs(j - i);
            }
        }
        return ans;
    }
};

  • class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        n = len(boxes)
        ans = [0] * n
        cnt = 0
        for i in range(1, n):
            if boxes[i - 1] == '1':
                cnt += 1
            ans[i] = ans[i - 1] + cnt
        cnt = s = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            s += cnt
            ans[i] += s
        return ans

############

class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        N = len(boxes)
        res = [0] * N
        for i in range(N):
            left = [i - j for j in range(i) if boxes[j] == "1"]
            right = [j - i for j in range(i + 1, N) if boxes[j] == "1"]
            res[i] = sum(left) + sum(right)
        return res

  • func minOperations(boxes string) []int {
	n := len(boxes)
	ans := make([]int, n)
	for i, cnt := 1, 0; i < n; i++ {
		if boxes[i-1] == '1' {
			cnt++
		}
		ans[i] = ans[i-1] + cnt
	}
	for i, cnt, s := n-2, 0, 0; i >= 0; i-- {
		if boxes[i+1] == '1' {
			cnt++
		}
		s += cnt
		ans[i] += s
	}
	return ans
}

  • function minOperations(boxes: string): number[] {
    const n = boxes.length;
    const ans = new Array(n).fill(0);
    for (let i = 1, count = 0; i < n; i++) {
        if (boxes[i - 1] === '1') {
            count++;
        }
        ans[i] = ans[i - 1] + count;
    }
    for (let i = n - 2, count = 0, sum = 0; i >= 0; i--) {
        if (boxes[i + 1] === '1') {
            count++;
        }
        sum += count;
        ans[i] += sum;
    }
    return ans;
}


  • impl Solution {
    pub fn min_operations(boxes: String) -> Vec<i32> {
        let s = boxes.as_bytes();
        let n = s.len();
        let mut ans = vec![0; n];
        let mut count = 0;
        for i in 1..n {
            if s[i - 1] == b'1' {
                count += 1;
            }
            ans[i] = ans[i - 1] + count;
        }
        let mut sum = 0;
        count = 0;
        for i in (0..n - 1).rev() {
            if s[i + 1] == b'1' {
                count += 1;
            }
            sum += count;
            ans[i] += sum;
        }
        ans
    }
}

All Problems

All Solutions