Given height and width of N rectangles. The task is to find the size of the largest subset such that no pair of rectangles fit within each other. Note that if H1 ≤ H2 and W1 ≤ W2 then rectangle 1 fits inside rectangle 2.
Examples:
Input: arr[] = {{1, 3}, {2, 2}, {1, 3}}
Output: 2
The required sub-set is {{1, 3}, {2, 2}}
{1, 3} is included only once as it can fit in {1, 3}
Input: arr[] = {{1, 5}, {2, 4}, {1, 1}, {3, 3}}
Output: 3
Approach: The above problem can be solved using Dynamic Programming and sorting. Initially, we can sort the N pairs on the basis of heights. A recursive function can be written where there will be two states.
The first state being, if the present rectangle does not fit in the previous rectangle or the vice versa, then we call for the next state with the present rectangle being the previous rectangle and moving to the next rectangle.
dp[present][previous] = max(dp[present][previous], 1 + dp[present + 1][present])
If it does fit in, we call the next state with the previous rectangle and moving to the next rectangle.
dp[present][previous] = max(dp[present][previous], dp[present + 1][previous])
Memoization can be further used to avoid repetitively the same states being called.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define N 10int dp[N][N];// Recursive function to get the largest subsetint findLongest(pair<int, int> a[], int n, int present, int previous){ // Base case when it exceeds if (present == n) { return 0; } // If the state has been visited previously else if (previous != -1) { if (dp[present][previous] != -1) return dp[present][previous]; } // Initialize int ans = 0; // No elements in subset yet if (previous == -1) { // First state which includes current index ans = 1 + findLongest(a, n, present + 1, present); // Second state which does not include current index ans = max(ans, findLongest(a, n, present + 1, previous)); } else { int h1 = a[previous].first; int h2 = a[present].first; int w1 = a[previous].second; int w2 = a[present].second; // If the rectangle fits in, then do not include // the current index in subset if ((h1 <= h2 && w1 <= w2)) { ans = max(ans, findLongest(a, n, present + 1, previous)); } else { // First state which includes current index ans = 1 + findLongest(a, n, present + 1, present); // Second state which does not include current index ans = max(ans, findLongest(a, n, present + 1, previous)); } } return dp[present][previous] = ans;}// Function to get the largest subsetint getLongest(pair<int, int> a[], int n){ // Initialize the DP table with -1 memset(dp, -1, sizeof dp); // Sort the array sort(a, a + n); // Get the answer int ans = findLongest(a, n, 0, -1); return ans;}// Driver codeint main(){ // (height, width) pairs pair<int, int> a[] = { { 1, 5 }, { 2, 4 }, { 1, 1 }, { 3, 3 } }; int n = sizeof(a) / sizeof(a[0]); cout << getLongest(a, n); return 0;} |
Python3
# Python3 implementation of the approach# Recursive function to get the# largest subsetdef findLongest(a, n, present, previous): # Base case when it exceeds if present == n: return 0 # If the state has been visited # previously elif previous != -1: if dp[present][previous] != -1: return dp[present][previous] # Initialize ans = 0 # No elements in subset yet if previous == -1: # First state which includes # current index ans = 1 + findLongest(a, n, present + 1, present) # Second state which does not # include current index ans = max(ans, findLongest(a, n, present + 1, previous)) else: h1 = a[previous][0] h2 = a[present][0] w1 = a[previous][1] w2 = a[present][1] # If the rectangle fits in, then do # not include the current index in subset if h1 <= h2 and w1 <= w2: ans = max(ans, findLongest(a, n, present + 1, previous)) else: # First state which includes # current index ans = 1 + findLongest(a, n, present + 1, present) # Second state which does not # include current index ans = max(ans, findLongest(a, n, present + 1, previous)) dp[present][previous] = ans return ans# Function to get the largest subsetdef getLongest(a, n): # Sort the array a.sort() # Get the answer ans = findLongest(a, n, 0, -1) return ans# Driver codeif __name__ == "__main__": # (height, width) pairs a = [[1, 5], [2, 4], [1, 1], [3, 3]] N = 10 # Initialize the DP table with -1 dp = [[-1 for i in range(N)] for j in range(N)] n = len(a) print(getLongest(a, n))# This code is contributed# by Rituraj Jain |
Javascript
<script>// JavaScript implementation of the approachvar N = 10;var dp = Array.from(Array(N), ()=>Array(N).fill(-1));// Recursive function to get the largest subsetfunction findLongest(a, n, present, previous){ // Base case when it exceeds if (present == n) { return 0; } // If the state has been visited previously else if (previous != -1) { if (dp[present][previous] != -1) return dp[present][previous]; } // Initialize var ans = 0; // No elements in subset yet if (previous == -1) { // First state which includes current index ans = 1 + findLongest(a, n, present + 1, present); // Second state which does not include current index ans = Math.max(ans, findLongest(a, n, present + 1, previous)); } else { var h1 = a[previous][0]; var h2 = a[present][0]; var w1 = a[previous][1]; var w2 = a[present][1]; // If the rectangle fits in, then do not include // the current index in subset if ((h1 <= h2 && w1 <= w2)) { ans = Math.max(ans, findLongest(a, n, present + 1, previous)); } else { // First state which includes current index ans = 1 + findLongest(a, n, present + 1, present); // Second state which does not include current index ans = Math.max(ans, findLongest(a, n, present + 1, previous)); } } return dp[present][previous] = ans;}// Function to get the largest subsetfunction getLongest(a, n){ // Initialize the DP table with -1 dp = Array.from(Array(N), ()=>Array(N).fill(-1)); // Sort the array a.sort((a,b)=>a-b); // Get the answer var ans = findLongest(a, n, 0, -1); return ans;}// Driver code// (height, width) pairsvar a = [ [ 1, 5 ], [ 2, 4 ], [ 1, 1 ], [ 3, 3 ] ];var n = a.length;document.write( getLongest(a, n));</script> |
3
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