Given two positive integers h and w representing the height h and width w which forms a rectangle. Also, there are two arrays of integers horizontalCuts and verticalCuts where horizontalCuts[i] is the distance from the top of the rectangle to the ith horizontal cut and similarly, verticalCuts[j] is the distance from the left of the rectangle to the jth vertical cut. The task is to find the maximum area of the rectangle after you cut at each horizontal and vertical position provided in the arrays horizontalCuts and verticalCuts. Since the answer can be a huge number, return this modulo 10^9 + 7.
Examples :Â
max area = 6
Input: h = 6, w = 4, horizontalCuts = [2, 5], verticalCuts = [1, 3]
Output: 6
Explanation: The figure above represents the given rectangle. Red lines are the horizontal cuts and blue lines are vertical cuts. After the rectangle is cut, the green piece of rectangle has the maximum area.Input: h = 5, w = 4, horizontalCuts = [3, 1], verticalCuts = [1]
Output: 9
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Approach: The problem can be solved by observing that-
- The horizontalCuts if perpendicular to any VerticalCut, then all the vertical slices cross all the horizontalCuts.
- Next, the maximum area of the rectangle must be enclosed by at least one vertical and one horizontal cut.
From the above observation, it is clear that we need to find the maximum distance between two horizontal cuts and two vertical cuts respectively, and multiply them to find the area of the rectangle. Follow the steps below to solve the problem:
- Sort both horizontalCuts and verticalCuts lists.
- Initialize two variables, say MaxHorizontal and MaxVertical as horizontalCuts[0] and verticalCuts[0] respectively, as to consider the closest rectangles towards axis both horizontally and vertically which will store the maximum horizontal and vertical lengths of the rectangle respectively.
- Iterate in the range [1, horizontalCuts.size()-1] using the variable i and perform the following steps:
- Modify the value of MaxHorizontal as max(MaxHorizontal, horizontalCuts[i] – horizontalCuts[i-1]).
- Modify the value of MaxVertical as max(MaxVertical, verticalCuts[i] – verticalCuts[i-1]).
- Print MaxHorizontal*MaxVertical as the answer.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;Â
class Solution {public:    // Returns the maximum area of rectangle    // after Horizontal and Vertical Cuts    int maxArea(int h, int w, vector<int>& horizontalCuts,                vector<int>& verticalCuts)    {Â
        // Sort the two arrays        sort(horizontalCuts.begin(), horizontalCuts.end());        sort(verticalCuts.begin(), verticalCuts.end());Â
        // Insert the right bound h and w        // in their respective vectors        horizontalCuts.push_back(h);        verticalCuts.push_back(w);                   //Initialising both by first indexs,          //to consider first rectangle formed by          //respective horizontal and vertical cuts        int maxHorizontal = horizontalCuts[0];         int maxVertical = verticalCuts[0];Â
        // Find the maximum Horizontal Length possible        for (int i = 1; i < horizontalCuts.size(); i++) {            int diff                = horizontalCuts[i] - horizontalCuts[i - 1];            maxHorizontal = max(maxHorizontal, diff);        }Â
        // Find the maximum vertical Length possible        for (int i = 1; i < verticalCuts.size(); i++) {            int diff                = verticalCuts[i] - verticalCuts[i - 1];            maxVertical = max(maxVertical, diff);        }Â
        // Return the maximum area of rectangle        return (int)((long)maxHorizontal * maxVertical                     % mod);    }};Â
// Driver Codeint main(){    // Class Call    Solution ob;       // Given Input    vector<int> hc = { 2, 5 }, vc = { 1, 3 };    int h = 6, v = 4;    // Function Call    cout << (ob.maxArea(6, 4, hc, vc));    return 0;} |
Java
// Java program for above approachimport java.util.*;class GFG{    // Returns the maximum area of rectangle    // after Horizontal and Vertical Cuts    public static int maxArea(int h, int w, ArrayList<Integer> horizontalCuts,                    ArrayList<Integer> verticalCuts)    {        // Sort the two arrays        Collections.sort(horizontalCuts);        Collections.sort(verticalCuts);Â
        // Insert the right bound h and w        // in their respective vectors                     //if the set is empty, add 0 as default value          if(horizontalCuts.size() == 0){            horizontalCuts.add(0);          }        if(verticalCuts.size() == 0){              verticalCuts.add(0);        }        horizontalCuts.add(h);        verticalCuts.add(w);Â
        // int maxHorizontal = 0;        // int maxVertical = 0;                    int maxHorizontal = horizontalCuts.get(0);        int maxVertical = verticalCuts.get(0);Â
        // Find the maximum Horizontal Length possible        for (int i = 1; i < horizontalCuts.size(); i++) {            int diff                    = horizontalCuts.get(i) - horizontalCuts.get(i-1);            maxHorizontal = Math.max(maxHorizontal, diff);        }Â
        // Find the maximum vertical Length possible        for (int i = 1; i < verticalCuts.size(); i++) {            int diff                    = verticalCuts.get(i) - verticalCuts.get(i - 1);            maxVertical = Math.max(maxVertical, diff);        }Â
        // Return the maximum area of rectangle        return (int)((long)maxHorizontal * maxVertical);    }    // Driver Code    public static void main(String[] args)    {        // Given Input        ArrayList<Integer> hc = new ArrayList<>();        hc.add(3);                 ArrayList<Integer> vc = new ArrayList<>();        vc.add(3);        int h = 5, v = 4;               // Function Call        System.out.println(maxArea(6, 4, hc, vc));    }}Â
//This code is contributed by Piyush Anand. |
Python3
# python 3 Program for the above approachmod = 1000000007Â
# Returns the maximum area of rectangle# after Horizontal and Vertical Cutsdef maxArea(h, w, horizontalCuts,            verticalCuts):Â
    # Sort the two arrays    horizontalCuts.sort()    verticalCuts.sort()Â
    # Insert the right bound h and w    # in their respective vectors    horizontalCuts.append(h)    verticalCuts.append(w)Â
    maxHorizontal = 0    maxVertical = 0Â
    # Find the maximum Horizontal Length possible    for i in range(1, len(horizontalCuts)):Â
        diff = horizontalCuts[i] - horizontalCuts[i - 1]        maxHorizontal = max(maxHorizontal, diff)Â
    # Find the maximum vertical Length possible    for i in range(1,                   len(verticalCuts)):        diff = verticalCuts[i] - verticalCuts[i - 1]        maxVertical = max(maxVertical, diff)Â
    # Return the maximum area of rectangle    return (int)(maxHorizontal * maxVertical                 % mod)Â
Â
# Driver Codeif __name__ == "__main__":Â
    # Given Input    hc = [2, 5]    vc = [1, 3]    h = 6    v = 4         # Function Call    print(maxArea(6, 4, hc, vc))Â
    # This code is contributed by ukasp. |
C#
// C# Program for the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{    static int mod = 1000000007;         // Returns the maximum area of rectangle    // after Horizontal and Vertical Cuts    static int maxArea(int h, int w, List<int> horizontalCuts, List<int> verticalCuts)    {        // Sort the two arrays        horizontalCuts.Sort();        verticalCuts.Sort();              // Insert the right bound h and w        // in their respective vectors        horizontalCuts.Add(h);        verticalCuts.Add(w);              int maxHorizontal = 0;        int maxVertical = 0;              // Find the maximum Horizontal Length possible        for(int i = 1; i < horizontalCuts.Count; i++)        {                  int diff = horizontalCuts[i] - horizontalCuts[i - 1];            maxHorizontal = Math.Max(maxHorizontal, diff);        }              // Find the maximum vertical Length possible        for(int i = 1; i < verticalCuts.Count; i++)        {            int diff = verticalCuts[i] - verticalCuts[i - 1];            maxVertical = Math.Max(maxVertical, diff);        }              // Return the maximum area of rectangle        return (int)(maxHorizontal * maxVertical % mod);    }       static void Main ()  {    // Given Input    List<int> hc = new List<int>(new int[]{ 2, 5 });    List<int> vc = new List<int>(new int[]{ 1, 3 });         // Function Call    Console.WriteLine(maxArea(6, 4, hc, vc));  }}Â
// This code is contributed by suresh07. |
Javascript
<script>Â Â Â Â Â Â Â Â // JavaScript program for the above approachÂ
        const mod = 1e9 + 7;Â
        class Solution {Â
            // Returns the maximum area of rectangle            // after Horizontal and Vertical Cuts            maxArea(h, w, horizontalCuts, verticalCuts) {Â
                // Sort the two arrays                horizontalCuts.sort(function (a, b) { return a - b; })                verticalCuts.sort(function (a, b) { return a - b; })Â
                // Insert the right bound h and w                // in their respective vectors                horizontalCuts.push(h);                verticalCuts.push(w);Â
                let maxHorizontal = 0;                let maxVertical = 0;Â
                // Find the maximum Horizontal Length possible                for (let i = 1; i < horizontalCuts.length; i++) {                    let diff                        = horizontalCuts[i] - horizontalCuts[i - 1];                    maxHorizontal = Math.max(maxHorizontal, diff);                }Â
                // Find the maximum vertical Length possible                for (let i = 1; i < verticalCuts.length; i++) {                    let diff                        = verticalCuts[i] - verticalCuts[i - 1];                    maxVertical = Math.max(maxVertical, diff);                }Â
                // Return the maximum area of rectangle                return parseInt(maxHorizontal * maxVertical                    % mod);            }        }Â
        // Driver CodeÂ
        // Class Call        let ob = new Solution();Â
        // Given Input        let hc = [2, 5], vc = [1, 3];        let h = 6, v = 4;        // Function Call        document.write(ob.maxArea(6, 4, hc, vc));Â
    // This code is contributed by Potta LokeshÂ
    </script> |
Output
6
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
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