PROBLEM DESCRIPTION
Given an integer n denoting the Length of a line segment. You need to cut the line segment in such a way that the cut length of a line segment each time is either x , y or z. Here x, y, and z are integers. After performing all the cut operations, your total number of cut segments must be maximum. Return the maximum number of cut segments possible.
Note: if no segment can be cut then return 0.
SOLUTION
f(i) represents the maximum number of cuts for a line segment of length i.
The possible lengths to cut the segment into are x, y, and z.
Option 1:
f(i - x) + 1Cut by
xlength. The maximum number of cuts for the remaining length will bef(i - x). Add 1 because we have made one additional cut.- Similarly, Option 2:
f(i - y) + 1 - And Option 3:
f(i - z) + 1
Since we are looking for the maximum number of cuts, we take the maximum of these three options.
This problem can be solved using a bottom-up approach with a dp[n + 1] array.
It is important to handle the “not possible” scenario: for example, when we attempt to cut the segment by x length, but the remaining length i - x cannot be further divided. We handle this by initializing all values in the dp array to -1. If dp[i - x] is -1, it means that it is not possible to divide the segment of length i - x using the available lengths x, y, and z.
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class Solution
{
public int maximizeCuts(int n, int x, int y, int z)
{
// dp array will store the maximum number of cuts possible for each length from 0 to n
int[] dp = new int[n+1];
// IMPORTANT: Initialize the dp array with -1
// -1 will indicate that it is not possible to divide the length i
// Take this test case to understand better:
// n = 7
// [5, 5, 2]
Arrays.fill(dp, -1);
// Base case: If the length of the segment is 0, no cuts are needed, so the number of cuts is 0
dp[0] = 0;
// Iterate through each length from 1 to n
for(int i = 1; i <= n; i++) {
// Check if it's possible to cut a segment of length `x` from `i`
// If dp[i-x] is not -1, that means it's possible to cut the remaining length (i-x),
// so we calculate the new number of cuts and update dp[i] if it results in more cuts
if(i >= x && dp[i - x] != -1) {
dp[i] = Math.max(dp[i], dp[i - x] + 1);
}
// Similarly, check if it's possible to cut a segment of length `y` from `i`
if(i >= y && dp[i - y] != -1) {
dp[i] = Math.max(dp[i], dp[i - y] + 1);
}
// Similarly, check if it's possible to cut a segment of length `z` from `i`
if(i >= z && dp[i - z] != -1) {
dp[i] = Math.max(dp[i], dp[i - z] + 1);
}
}
// If dp[n] is still -1, it means it's not possible to make any cuts, so return 0
// Otherwise, return the maximum number of cuts possible for length n
return dp[n] == -1 ? 0 : dp[n];
}
}