## Introduction:

In this article, we will discuss the Maximizing Line Segment Cuts algorithm. Given a line segment of length N and three possible cut lengths (x, y, and z), our goal is to maximize the total number of cut segments. We will explain the approach and provide a step-by-step algorithm to solve this problem efficiently. Let's dive in!

## Problem Statement:

Given a line segment of length N, we need to cut it in such a way that the cut length each time is either x, y, or z. Our objective is to maximize the total number of cut segments.

## Approach:

To solve this problem, we can use a dynamic programming approach. We'll create an array, dp[], where dp[i] represents the maximum number of cuts possible for a line segment of length i. Initially, all values in dp[] are set to -1, indicating that the corresponding length is not possible.

## Algorithm:

1. Initialize the dp[] array with -1 values and set dp = 0, as the number of cuts for a length of 0 is 0.
2. Iterate through each length from 0 to N:

- If dp[i] is -1, skip the current length as it is not possible to cut.

- If cutting a segment of length x is possible (i + x <= N), update dp[i + x] as max(dp[i + x], dp[i] + 1).

- If cutting a segment of length y is possible (i + y <= N), update dp[i + y] as max(dp[i + y], dp[i] + 1).

- If cutting a segment of length z is possible (i + z <= N), update dp[i + z] as max(dp[i + z], dp[i] + 1).

1. If dp[N] is still -1, set dp[N] = 0, as it means no segment can be cut.
2. Return dp[N] as the maximum number of cuts possible for the given line segment length N.

Let's understand the algorithm with the help of examples:

## Example 1:

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

dp[]: -1 -1 -1 -1 -1

0 -1 -1 -1 -1

Iterating through each length:

Length 0: dp = 0

Length 1: dp = max(dp, dp + 1) = max(-1, 0 + 1) = 1

Length 2: dp = max(dp, dp + 1) = max(-1, 0 + 1) = 1

Length 3: dp = max(dp, dp + 1) = max(-1, 1 + 1) = 2

Length 4: dp = max(dp, dp + 1) = max(-1, 1 + 1) = 2

The maximum number of cuts possible is 2.

## Example 2:

N = 5, x = 5, y = 3, z = 2

dp[]: -1 -1 -1 -1 -1 -1

0 -1 -1 -1 -1 -1

Iterating through each length:

Length 0: dp = 0

Length 1: dp = max(dp, dp + 1) = max(-1

, 0 + 1) = 1

Length 2: dp = max(dp, dp + 1) = max(-1, 0 + 1) = 1

Length 3: dp = max(dp, dp + 1) = max(-1, 0 + 1) = 1

Length 4: dp = max(dp, dp + 1) = max(-1, 0 + 1) = 1

Length 5: dp = max(dp, dp[5 - 5] + 1) = max(-1, 0 + 1) = 1

The maximum number of cuts possible is 1.

## Code:

#include <bits/stdc++.h>

using namespace std;

class Solution

{

public:

// Function to find the maximum number of cuts.

int maximizeTheCuts(int n, int x, int y, int z)

{

// Array to store the number of cuts at each length

int dp[n + 1];

// Initialize all values with -1

memset(dp, -1, sizeof(dp));

// If the length of the rod is 0, then total cuts will be 0.

// So, initialize dp with 0.

dp = 0;

for (int i = 0; i <= n; i++)

{

// If the certain length is not possible, skip it.

if (dp[i] == -1)

continue;

// If a segment of x is possible, update the number of cuts at the next length.

if (i + x <= n)

dp[i + x] = max(dp[i + x], dp[i] + 1);

// If a segment of y is possible, update the number of cuts at the next length.

if (i + y <= n)

dp[i + y] = max(dp[i + y], dp[i] + 1);

// If a segment of z is possible, update the number of cuts at the next length.

if (i + z <= n)

dp[i + z] = max(dp[i + z], dp[i] + 1);

}

// If no segment can be cut, return 0.

if (dp[n] == -1)

dp[n] = 0;

// Return the number of cuts corresponding to length n.

return dp[n];

}

};

int main()

{

// Taking the length of the line segment

int n;

cin >> n;

// Taking the types of segments

int x, y, z;

cin >> x >> y >> z;

Solution obj;

// Calling the maximizeTheCuts() function

cout << obj.maximizeTheCuts(n, x, y, z) << endl;

return 0;

}

## Conclusion:

In this article, we explored the Maximizing Line Segment Cuts algorithm. By using dynamic programming, we can efficiently determine the maximum number of cuts possible for a given line segment length. We discussed the approach, provided a step-by-step algorithm, and illustrated it with examples. This algorithm allows us to find an optimal solution and can be applied to similar problems involving maximizing the number of cuts.