Given a paper of size A x B. Task is to cut the paper into squares of any size. Find the minimum number of squares that can be cut from the paper.
Examples:
Input : 36 x 30 Output : 5 Explanation : 3 (squares of size 12x12) + 2 (squares of size 18x18) Input : 4 x 5 Output : 5 Explanation : 1 (squares of size 4x4) + 4 (squares of size 1x1)
Asked in : Google
We have already discussed the Greedy approach to solve this problem in previous article. But the previous approach doesn’t always work. For example, it fails for the above first test case. So, in this article we solve this problem using Dynamic Programming.
We know that if we want to cut minimum number of squares from the paper then we would have to cut largest square possible from the paper first and largest square will have same side as smaller side of the paper. For example if paper have the size 13 x 29, then maximum square will be of side 13. so we can cut 2 square of size 13 x 13 (29/13 = 2). Now remaining paper will have size 3 x 13. Similarly we can cut remaining paper by using 4 squares of size 3 x 3 and 3 squares of 1 x 1. So minimum 9 squares can be cut from the Paper of size 13 x 29.
Explanation:
minimumSquare is a function which tries to split the rectangle at some position. The function is called recursively for both parts. Try all possible splits and take the one with minimum result. The base case is when both sides are equal i.e the input is already a square, in which case the result is We are trying to find the cut which is nearest to the center which will lead us to our minimum result.
Assuming we have a rectangle with width is N and height is M.
- if (N == M), so it is a square and nothing need to be done.
- Otherwise, we can divide the rectangle into two other smaller one (N – x, M) and (x, M), so it can be solved recursively.
- Similarly, we can also divide it into (N, M – x) and (N, x)
Also we need to be aware of an edge case here which is N=11 and M=13 or vice versa. The following will be the best possible answer for the given test case:
Our Approach will return 8 for this case but as you can see we can cut the paper in 6 squares which is minimum. This happens because there is no vertical or horizontal line that cuts through the whole square in the optimum solution.
Below is the implementation of the above idea using Dynamic Programming.
C++
// C++ program to find minimum number of squares// to cut a paper using Dynamic Programming#include<bits/stdc++.h>using namespace std;const int MAX = 300;int dp[MAX][MAX];// Returns min number of squares neededint minimumSquare(int m, int n){ // Initializing max values to vertical_min // and horizontal_min int vertical_min = INT_MAX; int horizontal_min = INT_MAX; // N=11 & M=13 is a special case if(n==13 && m==11) return 6; if(m==13 && n==11) return 6; // If the given rectangle is already a square if (m == n) return 1; // If the answer for the given rectangle is // previously calculated return that answer if (dp[m][n]) return dp[m][n]; /* The rectangle is cut horizontally and vertically into two parts and the cut with minimum value is found for every recursive call. */ for (int i = 1;i<= m/2;i++) { // Calculating the minimum answer for the // rectangles with width equal to n and length // less than m for finding the cut point for // the minimum answer horizontal_min = min(minimumSquare(i, n) + minimumSquare(m-i, n), horizontal_min); } for (int j = 1;j<= n/2;j++) { // Calculating the minimum answer for the // rectangles with width less than n and // length equal to m for finding the cut // point for the minimum answer vertical_min = min(minimumSquare(m, j) + minimumSquare(m, n-j), vertical_min); } // Minimum of the vertical cut or horizontal // cut to form a square is the answer dp[m][n] = min(vertical_min, horizontal_min); return dp[m][n];}// Driver codeint main(){ int m = 30, n = 35; // Function call cout << minimumSquare(m, n); return 0;}// This code is contributed by Utkarsh Gupta |
Java
// Java program to find minimum number of// squares to cut a paper using Dynamic// Programmingimport java.io.*;class GFG { static int dp[][] = new int[300][300]; // Returns min number of squares needed static int minimumSquare(int m, int n) { // Initializing max values to // vertical_min and horizontal_min int vertical_min = Integer.MAX_VALUE; int horizontal_min = Integer.MAX_VALUE; // N=11 & M=13 is a special case if(n==13 && m==11) return 6; if(m==13 && n==11) return 6; // If the given rectangle is // already a square if (m == n) return 1; // If the answer for the given // rectangle is previously // calculated return that answer if (dp[m][n] != 0) return dp[m][n]; /* The rectangle is cut horizontally and vertically into two parts and the cut with minimum value is found for every recursive call. */ for (int i = 1; i <= m / 2; i++) { // Calculating the minimum answer // for the rectangles with width // equal to n and length less than // m for finding the cut point for // the minimum answer horizontal_min = Math.min(minimumSquare(i, n) + minimumSquare(m - i, n), horizontal_min); } for (int j = 1; j <= n / 2; j++) { // Calculating the minimum answer // for the rectangles with width // less than n and length equal to // m for finding the cut point for // the minimum answer vertical_min = Math.min(minimumSquare(m, j) + minimumSquare(m, n - j), vertical_min); } // Minimum of the vertical cut or // horizontal cut to form a square // is the answer dp[m][n] = Math.min(vertical_min, horizontal_min); return dp[m][n]; } // Driver code public static void main(String[] args) { int m = 30, n = 35; // Function call System.out.println(minimumSquare(m, n)); }}// This code is contributed by Prerna Saini |
Python3
# Python3 program to find minimum# number of squares# to cut a paper using Dynamic ProgrammingMAX = 300dp = [[0 for i in range(MAX)] for i in range(MAX)]# Returns min number of squares neededdef minimumSquare(m, n): # Initializing max values to # vertical_min # and horizontal_min vertical_min = 10000000000 horizontal_min = 10000000000 # N=11 & M=13 is a special case if n == 13 and m == 11: return 6 if m == 13 and n == 11: return 6 # If the given rectangle is # already a square if m == n: return 1 # If the answer for the given rectangle is # previously calculated return that answer if dp[m][n] != 0: return dp[m][n] # The rectangle is cut horizontally and # vertically into two parts and the cut # with minimum value is found for every # recursive call. for i in range(1, m//2+1): # Calculating the minimum answer for the # rectangles with width equal to n and length # less than m for finding the cut point for # the minimum answer horizontal_min = min(minimumSquare(i, n) + minimumSquare(m-i, n), horizontal_min) for j in range(1, n//2+1): # Calculating the minimum answer for the # rectangles with width equal to n and length # less than m for finding the cut point for # the minimum answer vertical_min = min(minimumSquare(m, j) + minimumSquare(m, n-j), vertical_min) # Minimum of the vertical cut or horizontal # cut to form a square is the answer dp[m][n] = min(vertical_min, horizontal_min) return dp[m][n]# Driver codeif __name__ == '__main__': m = 30 n = 35 # Function call print(minimumSquare(m, n))# This code is contributed by sahilshelangia |
C#
// C# program to find minimum number of// squares to cut a paper using Dynamic// Programmingusing System;class GFG { static int[, ] dp = new int[300, 300]; // Returns min number of squares needed static int minimumSquare(int m, int n) { // Initializing max values to // vertical_min and horizontal_min int vertical_min = int.MaxValue; int horizontal_min = int.MaxValue; // N=11 & M=13 is a special case if(n==13 && m==11) return 6; if(m==13 && n==11) return 6; // If the given rectangle is // already a square if (m == n) return 1; // If the answer for the given // rectangle is previously // calculated return that answer if (dp[m, n] != 0) return dp[m, n]; /* The rectangle is cut horizontally and vertically into two parts and the cut with minimum value is found for every recursive call. */ for (int i = 1; i <= m / 2; i++) { // Calculating the minimum answer // for the rectangles with width // equal to n and length less than // m for finding the cut point for // the minimum answer horizontal_min = Math.Min(minimumSquare(i, n) + minimumSquare(m - i, n), horizontal_min); } for (int j = 1; j <= n / 2; j++) { // Calculating the minimum answer // for the rectangles with width // less than n and length equal to // m for finding the cut point for // the minimum answer vertical_min = Math.Min(minimumSquare(m, j) + minimumSquare(m, n - j), vertical_min); } // Minimum of the vertical cut or // horizontal cut to form a square // is the answer dp[m, n] = Math.Min(vertical_min, horizontal_min); return dp[m, n]; } // Driver code public static void Main() { int m = 30, n = 35; // Function call Console.WriteLine(minimumSquare(m, n)); }}// This code is contributed by anuj_67. |
Javascript
<script> // Javascript program to find minimum number of // squares to cut a paper using Dynamic // Programming let dp = new Array(300); for(let i = 0; i < 300; i++) { dp[i] = new Array(300); for(let j = 0; j < 300; j++) { dp[i][j] = 0; } } // Returns min number of squares needed function minimumSquare(m, n) { // Initializing max values to // vertical_min and horizontal_min let vertical_min = Number.MAX_VALUE; let horizontal_min = Number.MAX_VALUE; // N=11 & M=13 is a special case if(n==13 && m==11) return 6; if(m==13 && n==11) return 6; // If the given rectangle is // already a square if (m == n) return 1; // If the answer for the given // rectangle is previously // calculated return that answer if (dp[m][n] != 0) return dp[m][n]; /* The rectangle is cut horizontally and vertically into two parts and the cut with minimum value is found for every recursive call. */ for (let i = 1; i <= parseInt(m / 2, 10); i++) { // Calculating the minimum answer // for the rectangles with width // equal to n and length less than // m for finding the cut point for // the minimum answer horizontal_min = Math.min(minimumSquare(i, n) + minimumSquare(m - i, n), horizontal_min); } for (let j = 1; j <= parseInt(n / 2, 10); j++) { // Calculating the minimum answer // for the rectangles with width // less than n and length equal to // m for finding the cut point for // the minimum answer vertical_min = Math.min(minimumSquare(m, j) + minimumSquare(m, n - j), vertical_min); } // Minimum of the vertical cut or // horizontal cut to form a square // is the answer dp[m][n] = Math.min(vertical_min, horizontal_min); return dp[m][n]; } let m = 30, n = 35; // Function call document.write(minimumSquare(m, n)); // This code is contributed by divyesh072019.</script> |
5
Time Complexity: O(N * M * (N + M))
Space Complexity: O(N * M)
This article is contributed by Ayush Govil, Aditya Nihal Kumar Singh and Deepanshu Aggarwal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
