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Sum of Areas of Rectangles possible for an array

Last Updated : 23 Feb, 2022
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Given an array, the task is to compute the sum of all possible maximum area rectangles which can be formed from the array elements. Also, you can reduce the elements of the array by at most 1. 

Examples: 

Input: a = {10, 10, 10, 10, 11,              10, 11, 10} Output: 210 Explanation:  We can form two rectangles one square (10 * 10)  and one (11 * 10). Hence, total area = 100 + 110 = 210.  Input: a = { 3, 4, 5, 6 } Output: 15 Explanation:  We can reduce 4 to 3 and 6 to 5 so that we got  rectangle of (3 * 5). Hence area = 15.  Input: a = { 3, 2, 5, 2 } Output: 0

Naive Approach: Check for all possible four elements of the array and then whichever can form a rectangle. In these rectangles, separate all those rectangles which are of the maximum area formed by these elements. After getting the rectangles and their areas, sum them all to get our desired solution.

Efficient Approach: To get the maximum area rectangle, first sort the elements of the array in the non-increasing array. After sorting, start the procedure to select the elements of the array. Here, selection of two elements of array (as length of rectangle) is possible if elements of array are equal (a[i] == a[i+1]) or if length of smaller element a[i+1] is one less than a[i] (in this case we have our length a[i+1] because a[i] is decreased by 1). One flag variable is maintained to check that whether we get length and breadth both. After getting the length, set the flag variable, now calculate the breadth in the same way as we have done for length, and sum the area of the rectangle. After getting length and breadth both, again set the flag variable false so that we will now search for a new rectangle. This process is repeated and lastly, the final sum of the area is returned. 

C++




// CPP code to find sum of all
// area rectangle possible
#include <bits/stdc++.h>
using namespace std;
 
// Function to find
// area of rectangles
int MaxTotalRectangleArea(int a[],
                          int n)
{
    // sorting the array in
    // descending order
    sort(a, a + n, greater<int>());
 
    // store the final sum of
    // all the rectangles area
    // possible
    int sum = 0;
    bool flag = false;
 
    // temporary variable to store
    // the length of rectangle
    int len;
     
    for (int i = 0; i < n; i++)
    {
         
        // Selecting the length of
        // rectangle so that difference
        // between any two number is 1
        // only. Here length is selected
        // so flag is set
        if ((a[i] == a[i + 1] || a[i] -
            a[i + 1] == 1) && (!flag))
        {
            // flag is set means
            // we have got length of
            // rectangle
            flag = true;
 
            // length is set to
            // a[i+1] so that if
            // a[i] a[i+1] is less
            // than by 1 then also
            // we have the correct
            // choice for length
            len = a[i + 1];
 
            // incrementing the counter
            // one time more as we have
            // considered a[i+1] element
            // also so.
            i++;
        }
 
        // Selecting the width of rectangle
        // so that difference between any
        // two number is 1 only. Here width
        // is selected so now flag is again
        // unset for next rectangle
        else if ((a[i] == a[i + 1] ||
                a[i] - a[i + 1] == 1) && (flag))
            {
                // area is calculated for
                // rectangle
                sum = sum + a[i + 1] * len;
 
                // flag is set false
                // for another rectangle
                // which we can get from
                // elements in array
                flag = false;
 
                // incrementing the counter
                // one time more as we have
                // considered a[i+1] element
                // also so.
                i++;
            }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int a[] = { 10, 10, 10, 10,
                11, 10, 11, 10,
                9, 9, 8, 8 };
    int n = sizeof(a) / sizeof(a[0]);
     
    cout << MaxTotalRectangleArea(a, n);
     
    return 0;
}
 
 

Java




// Java code to find sum of
// all area rectangle possible
import java.io.*;
import java.util.Arrays;
import java.util.*;
 
class GFG
{
    // Function to find
    // area of rectangles
    static int MaxTotalRectangleArea(Integer []a,
                                     int n)
    {
         
        // sorting the array in
        // descending order
        Arrays.sort(a, Collections.reverseOrder());
     
        // store the final sum of
        // all the rectangles area
        // possible
        int sum = 0;
        boolean flag = false;
     
        // temporary variable to
        // store the length of rectangle
        int len = 0;
         
        for (int i = 0; i < n; i++)
        {
             
            // Selecting the length of
            // rectangle so that difference
            // between any two number is 1
            // only. Here length is selected
            // so flag is set
            if ((a[i] == a[i + 1] ||
                 a[i] - a[i+1] == 1) &&
                               !flag)
            {
                // flag is set means
                // we have got length of
                // rectangle
                flag = true;
     
                // length is set to
                // a[i+1] so that if
                // a[i] a[i+1] is less
                // than by 1 then also
                // we have the correct
                // choice for length
                len = a[i+1];
     
                // incrementing the counter
                // one time more as we have
                // considered a[i+1] element
                // also so.
                i++;
            }
     
            // Selecting the width of rectangle
            // so that difference between any
            // two number is 1 only. Here width
            // is selected so now flag is again
            // unset for next rectangle
            else if ((a[i] == a[i + 1] ||
                      a[i] - a[i+1] == 1) &&
                                    (flag))
                {
                    // area is calculated for
                    // rectangle
                    sum = sum + a[i+1] * len;
     
                    // flag is set false
                    // for another rectangle
                    // which we can get from
                    // elements in array
                    flag = false;
     
                    // incrementing the counter
                    // one time more as we have
                    // considered a[i+1] element
                    // also so.
                    i++;
                }
        }
     
        return sum;
    }
     
    // Driver code
    public static void main (String args[])
    {
    Integer []a = { 10, 10, 10, 10,
                11, 10, 11, 10,
                9, 9, 8, 8 };
    int n = a.length;
     
    System.out.print(MaxTotalRectangleArea(a, n));
    }
}
// This code is contributed by
// Manish Shaw(manishshaw1)
 
 

Python3




# Python3 code to find sum
# of all area rectangle
# possible
 
# Function to find
# area of rectangles
def MaxTotalRectangleArea(a, n) :
     
    # sorting the array in
    # descending order
    a.sort(reverse = True)
 
    # store the final sum of
    # all the rectangles area
    # possible
    sum = 0
    flag = False
 
    # temporary variable to store
    # the length of rectangle
    len = 0
    i = 0
     
    while (i < n-1) :
        if(i != 0) :
            i = i + 1
             
        # Selecting the length of
        # rectangle so that difference
        # between any two number is 1
        # only. Here length is selected
        # so flag is set
        if ((a[i] == a[i + 1] or
             a[i] - a[i + 1] == 1)
              and flag == False) :
                   
            # flag is set means
            # we have got length of
            # rectangle
            flag = True
 
            # length is set to
            # a[i+1] so that if
            # a[i+1] is less than a[i]
            # by 1 then also we have
            # the correct choice for length
            len = a[i + 1]
 
            # incrementing the counter
            # one time more as we have
            # considered a[i+1] element
            # also so.
            i = i + 1
 
        # Selecting the width of rectangle
        # so that difference between any
        # two number is 1 only. Here width
        # is selected so now flag is again
        # unset for next rectangle
        elif ((a[i] == a[i + 1] or
              a[i] - a[i + 1] == 1)
                and flag == True) :
                     
            # area is calculated for
            # rectangle
            sum = sum + a[i + 1] * len
             
            # flag is set false
            # for another rectangle
            # which we can get from
            # elements in array
            flag = False
 
            # incrementing the counter
            # one time more as we have
            # considered a[i+1] element
            # also so.
            i = i + 1
         
    return sum
 
# Driver code
a = [ 10, 10, 10, 10, 11, 10,
          11, 10, 9, 9, 8, 8 ]
n = len(a)
 
print (MaxTotalRectangleArea(a, n))
 
# This code is contributed by
# Manish Shaw (manishshaw1)
 
 

C#




// C# code to find sum of all area rectangle
// possible
using System;
 
class GFG {
 
    // Function to find
    // area of rectangles
    static int MaxTotalRectangleArea(int []a,
                                       int n)
    {
         
        // sorting the array in descending
        // order
        Array.Sort(a);
        Array.Reverse(a);
     
        // store the final sum of all the
        // rectangles area possible
        int sum = 0;
        bool flag = false;
     
        // temporary variable to store the
        // length of rectangle
        int len =0;
         
        for (int i = 0; i < n; i++)
        {
             
            // Selecting the length of
            // rectangle so that difference
            // between any two number is 1
            // only. Here length is selected
            // so flag is set
            if ((a[i] == a[i + 1] || a[i] -
                a[i + 1] == 1) && (!flag))
            {
                // flag is set means
                // we have got length of
                // rectangle
                flag = true;
     
                // length is set to
                // a[i+1] so that if
                // a[i] a[i+1] is less
                // than by 1 then also
                // we have the correct
                // choice for length
                len = a[i + 1];
     
                // incrementing the counter
                // one time more as we have
                // considered a[i+1] element
                // also so.
                i++;
            }
     
            // Selecting the width of rectangle
            // so that difference between any
            // two number is 1 only. Here width
            // is selected so now flag is again
            // unset for next rectangle
            else if ((a[i] == a[i + 1] ||
                    a[i] - a[i + 1] == 1) && (flag))
                {
                    // area is calculated for
                    // rectangle
                    sum = sum + a[i + 1] * len;
     
                    // flag is set false
                    // for another rectangle
                    // which we can get from
                    // elements in array
                    flag = false;
     
                    // incrementing the counter
                    // one time more as we have
                    // considered a[i+1] element
                    // also so.
                    i++;
                }
        }
     
        return sum;
    }
     
    // Driver code
    static public void Main ()
    {
            int []a = { 10, 10, 10, 10,
                        11, 10, 11, 10,
                        9, 9, 8, 8 };
    int n = a.Length;
     
    Console.WriteLine(
                MaxTotalRectangleArea(a, n));
    }
}
 
// This code is contributed by anuj_67.
 
 

PHP




<?php
// PHP code to find sum
// of all area rectangle
// possible
 
// Function to find
// area of rectangles
function MaxTotalRectangleArea( $a, $n)
{
    // sorting the array in
    // descending order
    rsort($a);
 
    // store the final sum of
    // all the rectangles area
    // possible
    $sum = 0;
    $flag = false;
 
    // temporary variable to store
    // the length of rectangle
    $len;
     
    for ( $i = 0; $i < $n; $i++)
    {
         
        // Selecting the length of
        // rectangle so that difference
        // between any two number is 1
        // only. Here length is selected
        // so flag is set
        if (($a[$i] == $a[$i + 1] or $a[$i] -
             $a[$i + 1] == 1) and (!$flag))
        {
            // flag is set means
            // we have got length of
            // rectangle
            $flag = true;
 
            // length is set to
            // a[i+1] so that if
            // a[i+1] is less than a[i]
            // by 1 then also we have
            // the correct choice for length
            $len = $a[$i + 1];
 
            // incrementing the counter
            // one time more as we have
            // considered a[i+1] element
            // also so.
            $i++;
        }
 
        // Selecting the width of rectangle
        // so that difference between any
        // two number is 1 only. Here width
        // is selected so now flag is again
        // unset for next rectangle
        else if (($a[$i] == $a[$i + 1] or
                  $a[$i] - $a[$i + 1] == 1) and
                 ($flag))
            {
                // area is calculated for
                // rectangle
                $sum = $sum + $a[$i + 1] * $len;
 
                // flag is set false
                // for another rectangle
                // which we can get from
                // elements in array
                $flag = false;
 
                // incrementing the counter
                // one time more as we have
                // considered a[i+1] element
                // also so.
                $i++;
            }
    }
 
    return $sum;
}
 
// Driver code
$a = array( 10, 10, 10, 10,
            11, 10, 11, 10,
            9, 9, 8, 8 );
$n = count($a);
 
echo MaxTotalRectangleArea($a, $n);
 
//This code is contributed by anuj_67.
?>
 
 

Javascript




<script>
 
// Javascript code to find sum of all
// area rectangle possible
 
// Function to find
// area of rectangles
function MaxTotalRectangleArea( a, n)
{
    // sorting the array in
    // descending order
    a.sort();
    a.reverse();
 
    // store the final sum of
    // all the rectangles area
    // possible
    let sum = 0;
    let flag = false;
 
    // temporary variable to store
    // the length of rectangle
    let len;
     
    for (let i = 0; i < n; i++)
    {
         
        // Selecting the length of
        // rectangle so that difference
        // between any two number is 1
        // only. Here length is selected
        // so flag is set
        if ((a[i] == a[i + 1] || a[i] -
            a[i + 1] == 1) && (!flag))
        {
            // flag is set means
            // we have got length of
            // rectangle
            flag = true;
 
            // length is set to
            // a[i+1] so that if
            // a[i] a[i+1] is less
            // than by 1 then also
            // we have the correct
            // choice for length
            len = a[i + 1];
 
            // incrementing the counter
            // one time more as we have
            // considered a[i+1] element
            // also so.
            i++;
        }
 
        // Selecting the width of rectangle
        // so that difference between any
        // two number is 1 only. Here width
        // is selected so now flag is again
        // unset for next rectangle
        else if ((a[i] == a[i + 1] ||
                a[i] - a[i + 1] == 1) && (flag))
            {
                // area is calculated for
                // rectangle
                sum = sum + a[i + 1] * len;
 
                // flag is set false
                // for another rectangle
                // which we can get from
                // elements in array
                flag = false;
 
                // incrementing the counter
                // one time more as we have
                // considered a[i+1] element
                // also so.
                i++;
            }
    }
 
    return sum;
}
 
 
// Driver Code
 
let a = [ 10, 10, 10, 10,
        11, 10, 11, 10,
        9, 9, 8, 8 ];
let n = a.length;
     
document.write(MaxTotalRectangleArea(a, n));
 
</script>
 
 
Output
282

Time Complexity: O(nlog(n)) 
Auxiliary Space: O(1)
 



Next Article
Largest lexicographic array with at-most K consecutive swaps

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Article Tags :
  • DSA
  • Geometric
  • Greedy
  • Mathematical
  • Sorting
  • area-volume-programs
  • square-rectangle
Practice Tags :
  • Geometric
  • Greedy
  • Mathematical
  • Sorting

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      Given an array a, your task is to convert it into a non-increasing form such that we can either increment or decrement the array value by 1 in the minimum changes possible. Examples : Input : a[] = {3, 1, 2, 1}Output : 1Explanation : We can convert the array into 3 1 1 1 by changing 3rd element of a
      11 min read

    • Sorting array with reverse around middle
      Consider the given array arr[], we need to find if we can sort array with the given operation. The operation is We have to select a subarray from the given array such that the middle element(or elements (in case of even number of elements)) of subarray is also the middle element(or elements (in case
      6 min read

    • Sum of Areas of Rectangles possible for an array
      Given an array, the task is to compute the sum of all possible maximum area rectangles which can be formed from the array elements. Also, you can reduce the elements of the array by at most 1. Examples: Input: a = {10, 10, 10, 10, 11, 10, 11, 10} Output: 210 Explanation: We can form two rectangles o
      13 min read

    • Largest lexicographic array with at-most K consecutive swaps
      Given an array arr[], find the lexicographically largest array that can be obtained by performing at-most k consecutive swaps. Examples : Input : arr[] = {3, 5, 4, 1, 2} k = 3 Output : 5, 4, 3, 2, 1 Explanation : Array given : 3 5 4 1 2 After swap 1 : 5 3 4 1 2 After swap 2 : 5 4 3 1 2 After swap 3
      9 min read

    • Partition into two subsets of lengths K and (N - k) such that the difference of sums is maximum
      Given an array of non-negative integers of length N and an integer K. Partition the given array into two subsets of length K and N - K so that the difference between the sum of both subsets is maximum. Examples : Input : arr[] = {8, 4, 5, 2, 10} k = 2 Output : 17 Explanation : Here, we can make firs
      7 min read

    Greedy algorithm on Operating System

    • Program for First Fit algorithm in Memory Management
      Prerequisite : Partition Allocation MethodsIn the first fit, the partition is allocated which is first sufficient from the top of Main Memory.Example : Input : blockSize[] = {100, 500, 200, 300, 600}; processSize[] = {212, 417, 112, 426};Output:Process No. Process Size Block no. 1 212 2 2 417 5 3 11
      8 min read

    • Program for Best Fit algorithm in Memory Management
      Prerequisite : Partition allocation methodsBest fit allocates the process to a partition which is the smallest sufficient partition among the free available partitions. Example: Input : blockSize[] = {100, 500, 200, 300, 600}; processSize[] = {212, 417, 112, 426}; Output: Process No. Process Size Bl
      8 min read

    • Program for Worst Fit algorithm in Memory Management
      Prerequisite : Partition allocation methodsWorst Fit allocates a process to the partition which is largest sufficient among the freely available partitions available in the main memory. If a large process comes at a later stage, then memory will not have space to accommodate it. Example: Input : blo
      8 min read

    • Program for Shortest Job First (or SJF) CPU Scheduling | Set 1 (Non- preemptive)
      The shortest job first (SJF) or shortest job next, is a scheduling policy that selects the waiting process with the smallest execution time to execute next. SJN, also known as Shortest Job Next (SJN), can be preemptive or non-preemptive. Characteristics of SJF Scheduling: Shortest Job first has the
      13 min read

    • Job Scheduling with two jobs allowed at a time
      Given a 2d array jobs[][] of order n * 2, where each element jobs[i], contains two integers, representing the start and end time of the job. Your task is to check if it is possible to complete all the jobs, provided that two jobs can be done simultaneously at a particular moment. Note: If a job star
      6 min read

    • Optimal Page Replacement Algorithm
      In operating systems, whenever a new page is referred and not present in memory, page fault occurs, and Operating System replaces one of the existing pages with newly needed page. Different page replacement algorithms suggest different ways to decide which page to replace. The target for all algorit
      3 min read

    Greedy algorithm on Graph

    • Prim’s Algorithm for Minimum Spanning Tree (MST)
      Prim’s algorithm is a Greedy algorithm like Kruskal's algorithm. This algorithm always starts with a single node and moves through several adjacent nodes, in order to explore all of the connected edges along the way. The algorithm starts with an empty spanning tree. The idea is to maintain two sets
      15+ min read

    • Boruvka's algorithm | Greedy Algo-9
      We have discussed the following topics on Minimum Spanning Tree.Applications of Minimum Spanning Tree Problem Kruskal’s Minimum Spanning Tree Algorithm Prim’s Minimum Spanning Tree AlgorithmIn this post, Boruvka's algorithm is discussed. Like Prim's and Kruskal's, Boruvka’s algorithm is also a Greed
      15+ min read

    • Dial's Algorithm (Optimized Dijkstra for small range weights)
      Dijkstra’s shortest path algorithm runs in O(Elog V) time when implemented with adjacency list representation (See C implementation and STL based C++ implementations for details). Input : Source = 0, Maximum Weight W = 14Output : Vertex Distance from Source 0 0 1 4 2 12 3 19 4 21 5 11 6 9 7 8 8 14 C
      15+ min read

    • Minimum cost to connect all cities
      There are n cities and there are roads in between some of the cities. Somehow all the roads are damaged simultaneously. We have to repair the roads to connect the cities again. There is a fixed cost to repair a particular road. Input is in the form of edges {u, v, w} where, u and v are city indices.
      7 min read

    • Number of single cycle components in an undirected graph
      Given a set of 'n' vertices and 'm' edges of an undirected simple graph (no parallel edges and no self-loop), find the number of single-cycle components present in the graph. A single-cyclic component is a graph of n nodes containing a single cycle through all nodes of the component. Example: Let us
      9 min read

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