16. Maximize The Cut Segments

The problem can be found at the following link: Question Link

Problem Description

Given an integer n denoting the length of a line segment, the task is to cut the line segment into the maximum number of smaller segments such that the length of each segment is either x, y, or z. Here, x, y, and z are integers representing possible segment lengths.

If it is not possible to make any cuts, return 0.

Example:

Input:

n = 4, x = 2, y = 1, z = 1

Output:

4

Explanation: The total length is 4, and the cut lengths are 2, 1, and 1. We can make a maximum of 4 segments, each of length 1.

My Approach

1. Dynamic Programming (DP) Initialization:

   • Create a DP array dp of size n + 1 initialized to -1, except dp[0] which is set to 0. This array will store the maximum number of segments that can be made for each length from 0 to n.

   • Initialize an array cuts containing the values x, y, and z.

2. Filling the DP Array:

   • Iterate through each length i from 1 to n.

   • For each length i, iterate over the possible cuts in cuts. If the current length i is greater than or equal to the cut length and the previous length i - cut can be achieved (dp[i - cut] != -1), update dp[i] as the maximum of its current value and dp[i - cut] + 1.

3. Final Result:

   • The final result is stored in dp[n]. If dp[n] is -1, return 0, indicating no segments can be made. Otherwise, return dp[n], which represents the maximum number of segments.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), as we are iterating through each possible length from 1 to n and checking up to three possible cuts for each length.

• Expected Auxiliary Space Complexity: O(n), as we are using a DP array of size n + 1 to store the maximum number of segments for each length.

Code (C++)

class Solution {
public:
    int maximizeTheCuts(int n, int x, int y, int z) {
        int dp[n + 1];
        memset(dp, -1, sizeof(dp));
        dp[0] = 0;

        int cuts[] = {x, y, z};

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < 3; j++) {
                if (i >= cuts[j] && dp[i - cuts[j]] != -1) {
                    dp[i] = max(dp[i], dp[i - cuts[j]] + 1);
                }
            }
        }

        return dp[n] == -1 ? 0 : dp[n];
    }
};

Code (Java)

class Solution {
    public int maximizeCuts(int n, int x, int y, int z) {
        int[] dp = new int[n + 1];
        Arrays.fill(dp, -1);
        dp[0] = 0;

        int[] cuts = {x, y, z};

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < 3; j++) {
                if (i >= cuts[j] && dp[i - cuts[j]] != -1) {
                    dp[i] = Math.max(dp[i], dp[i - cuts[j]] + 1);
                }
            }
        }

        return dp[n] == -1 ? 0 : dp[n];
    }
}

Code (Python)

class Solution:
    def maximizeTheCuts(self, n, x, y, z):
        dp = [-1] * (n + 1)
        dp[0] = 0

        cuts = [x, y, z]

        for i in range(1, n + 1):
            for cut in cuts:
                if i >= cut and dp[i - cut] != -1:
                    dp[i] = max(dp[i], dp[i - cut] + 1)

        return dp[n] if dp[n] != -1 else 0

Contribution and Support

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