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For all Array elements find Product of Sum of all smaller and Sum of all greater elements

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  • Difficulty Level : Easy
  • Last Updated : 27 Jul, 2022
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Given an array arr[] of integers of length N, the task is to find the product of the sum of all the numbers larger than that number with the sum of all the numbers less than that number for each number in the array.

Examples

Input: arr[] = {8, 4, 9, 3}, N = 4
Output:- 63, 51, 0, 0
Explanation:
For first number 8: Sum of elements smaller than this is (4 + 3) = 7
and sum of elements greater than this is 9. hence output is 7*9 = 63.  
For 4: Summation of elements smaller than this is 3 and 
summation of elements greater than this is (9 + 8) = 17 hence output is 17*3 = 51
For 9: Summation of elements smaller than this is (8 + 4 + 3 ) = 15 and
summation of elements greater than this is 0. Hence output is 15*0 = 0
For 3: Summation of elements smaller than this is 0 and 
summation of elements greater than this is (8 + 4 + 9) = 21. Hence output is 0*21 = 0

Input: arr[] = {1, 4, 7, 3, 6}, N = 5
Output: 0, 52, 0, 17, 56

 

Approach: The idea is to find the sum of all the smaller elements and all the greater elements for each array element and then find the product for those.

Follow the steps below to solve this problem:

  •  Iterate through all the elements of an array.
    • For each element of an array keep adding elements smaller than the current and elements greater than the current element in two different variables. 
    • Multiply these two variables ( i.e  which are storing elements greater and less than the current element ) 
    • Store the answer for each element.
  • Return the array storing the answers.

Below is the implementation of the above approach: 

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the answer
vector<int> constructArray(int arr[], int n)
{
    int i, s = 0, j = 0, sum = 0;
    vector<int> v;
    for (i = 0; i < n; i++) {
        s = 0, sum = 0;
        for (j = 0; j < n; j++) {
            if (arr[j] != arr[i]) {
 
                // Addition of elements
                // greater than ith indexed
                // element
                if (arr[i] > arr[j])
                    s += arr[j];
 
                // Addition of elements
                // less than ith indexed
                // element
                else
                    sum += arr[j];
            }
        }
 
        // Storing the product of elements
        // greater than ith indexed elements
        // with elements less than ith
        // indexed element.
        v.push_back(s * sum);
    }
    return v;
}
 
// Driver Code
int main()
{
    int N = 4;
    int arr[] = { 8, 4, 9, 3 };
 
    // Function call
    vector<int> ans = constructArray(arr, N);
    for (int x : ans)
        cout << x << " ";
    return 0;
}


C




#include <stdio.h>
#include <stdlib.h>
 
// Function to find the answer
void constructArray(int arr[], int n)
{
    int i, s = 0, j = 0, sum = 0;
    int idx=0;
    int v[n];
    for (i = 0; i < n; i++) {
        s = 0, sum = 0;
        for (j = 0; j < n; j++) {
            if (arr[j] != arr[i]) {
 
                // Addition of elements
                // greater than ith indexed
                // element
                if (arr[i] > arr[j])
                    s += arr[j];
 
                // Addition of elements
                // less than ith indexed
                // element
                else
                    sum += arr[j];
            }
        }
 
        // Storing the product of elements
        // greater than ith indexed elements
        // with elements less than ith
        // indexed element.
        v[idx++]=(s * sum);
    }
     
    for(int i=0;i<idx;i++){
      printf("%d , ",v[i]);
    }
     
}
 
// Driver Code
int main()
{
    int N = 4;
    int arr[] = { 8, 4, 9, 3 };
 
    // Function call
    constructArray(arr, N);
     
    return 0;
}


Java




// Java code to implement the approach
import java.io.*;
import java.util.*;
 
class GFG
{
   
  // Function to find the answer
  public static ArrayList<Integer>
    constructArray(int arr[], int n)
  {
    int i = 0;
    int s = 0;
    int j = 0;
    int sum = 0;
    ArrayList<Integer> v = new ArrayList<Integer>();
    for (i = 0; i < n; i++) {
      s = 0;
      sum = 0;
      for (j = 0; j < n; j++) {
        if (arr[j] != arr[i]) {
 
          // Addition of elements
          // greater than ith indexed
          // element
          if (arr[i] > arr[j])
            s += arr[j];
 
          // Addition of elements
          // less than ith indexed
          // element
          else
            sum += arr[j];
        }
      }
 
      // Storing the product of elements
      // greater than ith indexed elements
      // with elements less than ith
      // indexed element.
      v.add(s * sum);
    }
    return v;
  }
 
  public static void main(String[] args)
  {
    int N = 4;
    int arr[] = { 8, 4, 9, 3 };
 
    // Function call
    ArrayList<Integer> ans = constructArray(arr, N);
    for (Integer x : ans)
      System.out.print(x + " ");
  }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Python3 code to implement the approach
 
# Function to find the answer
def constructArray(arr, n):
    s = 0
    sums = 0
    v = []
    for i in range(n):
        s = 0
        sums = 0
        for j in range(n):
            if arr[j] != arr[i]:
                # Addition of elements
                # greater than ith indexed
                # element
                if arr[i] > arr[j]:
                    s += arr[j]
                # Addition of elements
                # less than ith indexed
                # element
                else:
                    sums += arr[j]
        # Storing the product of elements
        # greater than ith indexed elements
        # with elements less than ith
        # indexed element.
        v.append(s * sums)
 
    return v
 
# Driver Code
N = 4
arr = [8, 4, 9, 3]
 
# Function Call
ans = constructArray(arr, N)
print(" ".join(list(map(str, ans))))
 
# this code is contributed by phasing17


C#




// C# code to implement the approach
using System;
using System.Collections;
 
public class GFG{
 
  // Function to find the answer
  public static ArrayList
    constructArray(int[] arr, int n)
  {
    int i = 0;
    int s = 0;
    int j = 0;
    int sum = 0;
    ArrayList v = new ArrayList();
    for (i = 0; i < n; i++) {
      s = 0;
      sum = 0;
      for (j = 0; j < n; j++) {
        if (arr[j] != arr[i]) {
 
          // Addition of elements
          // greater than ith indexed
          // element
          if (arr[i] > arr[j])
            s += arr[j];
 
          // Addition of elements
          // less than ith indexed
          // element
          else
            sum += arr[j];
        }
      }
 
      // Storing the product of elements
      // greater than ith indexed elements
      // with elements less than ith
      // indexed element.
      v.Add(s * sum);
    }
    return v;
  }
  static public void Main (){
 
    int N = 4;
    int[] arr = { 8, 4, 9, 3 };
 
    // Function call
    ArrayList ans = constructArray(arr, N);
    foreach (var x in ans)
      Console.Write(x + " ");
  }
}
 
// This code is contributed by hrithikgarg03188.


Javascript




<script>
     // JavaScript code for the above approach
 
     // Function to find the answer
     function constructArray(arr, n) {
         let i, s = 0, j = 0, sum = 0;
         let v = [];
         for (i = 0; i < n; i++) {
             s = 0, sum = 0;
             for (j = 0; j < n; j++) {
                 if (arr[j] != arr[i]) {
 
                     // Addition of elements
                     // greater than ith indexed
                     // element
                     if (arr[i] > arr[j])
                         s += arr[j];
 
                     // Addition of elements
                     // less than ith indexed
                     // element
                     else
                         sum += arr[j];
                 }
             }
 
             // Storing the product of elements
             // greater than ith indexed elements
             // with elements less than ith
             // indexed element.
             v.push(s * sum);
         }
         return v;
     }
 
     // Driver Code
     let N = 4;
     let arr = [8, 4, 9, 3];
 
     // Function call
     let ans = constructArray(arr, N);
     for (let x of ans)
         document.write(x + " ")
 
 // This code is contributed by Potta Lokesh
 </script>


Output

63 51 0 0 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach:

Above solution works in O(N2) time, we can write a solution in O(n*Logn) time and with O(N) space complexity.

The idea is to sort the array and finding prefix_sum of sorted array.Now prefix sum will help us in finding the sum of all smaller elements  and sum of all larger elements of the current element.

Follow the steps below for understanding…

  • sort the given array and store resultant array into another array
  • find prefix_sum array of sorted array
  • Now find index of each element of given array into sorted array using binary search.
  • With the help of prefix array,we can directly calculate sum of all smaller element and sum of all larger element of the current element.
  • to calculate the resultant value corresponding to each element, do smaller*larger.

   finally return the array storing the answers

Below is the implementation of the above approach:

Java




import java.util.*;
import java.io.*;
 
// Java program for the above approach
class GFG{
 
  static int binary_search(ArrayList<Integer> arr, int low, int high, int x){
    // Check base case
    if (high >= low){
      int mid = (high + low); // 2
      // If element is present at the middle itself
      if (arr.get(mid) == x){
        return mid;
 
        // If element is smaller than mid, then it can only
        // be present in left subarray
      }else if(arr.get(mid) > x){
        return binary_search(arr, low, mid - 1, x);
        // Else the element can only be present in right subarray
 
      }else{
        return binary_search(arr, mid + 1, high, x);
      }
    }
    return -1;
  }
 
  static ArrayList<Integer> constructArray(ArrayList<Integer> arr, int n){
    // sorting the arr and assigning to another array
    ArrayList<Integer> arr2 = new ArrayList<Integer>(arr);
    Collections.sort(arr2);
 
    // Initializing prefix_sum array
    int[] prefix_sum = new int[n];
 
    // first element of prefix_sum and given array will be the same
    prefix_sum[0] = arr2.get(0);
 
    // iterating through 0 to n-1 and
    // calculating prefix_sum
    for (int i = 1 ; i < n ; i++){
      prefix_sum[i] = prefix_sum[i-1] + arr2.get(i);
    }
 
    for (int i = 0 ; i < n ; i++){
 
      // storing index of each element of given array
      // in our sorted array using binary search
      int element_index = binary_search(arr2, 0, n - 1, arr.get(i));
 
      // storing sum of all smaller elements into smaller
      int smaller = prefix_sum[element_index]-arr.get(i);
 
      // storing sum of all larger element into larger
      int larger = prefix_sum[n-1]-prefix_sum[element_index];
       
      // multiplying smaller and larger and '
      // storing into arr[i]'
      arr.set(i, smaller * larger);
    }
 
    return arr;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int N = 4;
    ArrayList<Integer> arr = new ArrayList<Integer>(
      List.of(
        8, 4, 9, 3
      )
    );
    // Function Call
    ArrayList<Integer> ans = constructArray(arr, N);
    for(int i = 0 ; i < ans.size() ; i++){
      System.out.print(ans.get(i) + " ");
    }
    System.out.println("");
  }
}
 
// This code is contributed by subhamgoyal2014.


Python3




def binary_search(arr, low, high, x):
    # Check base case
    if high >= low:
        mid = (high + low) // 2
        # If element is present at the middle itself
        if arr[mid] == x:
            return mid
        # If element is smaller than mid, then it can only
        # be present in left subarray
        elif arr[mid] > x:
            return binary_search(arr, low, mid - 1, x)
        # Else the element can only be present in right subarray
        else:
            return binary_search(arr, mid + 1, high, x)
 
 
def constructArray(arr, n):
    # sorting the arr and assigning to another array
    arr2 = sorted(arr)
    # Initializing prefix_sum array
    prefix_sum = [0]*n
    # first element of prefix_sum and given array will be the same
    prefix_sum[0] = arr2[0]
    # iterating through 0 to n-1 and
    # calculating prefix_sum
    for i in range(1, n):
        prefix_sum[i] = prefix_sum[i-1]+arr2[i]
    for i in range(n):
        # storing index of each element of given array
        # in our sorted array using binary search
        element_index = binary_search(arr2, 0, n-1, arr[i])
        # storing sum of all smaller elements into smaller
        smaller = prefix_sum[element_index]-arr[i]
        # storing sum of all larger element into larger
        larger = prefix_sum[n-1]-prefix_sum[element_index]
        # multiplying smaller and larger and '
        # storing into arr[i]'
        arr[i] = smaller*larger
 
    return arr
 
 
# Driver Code
N = 4
arr = [8, 4, 9, 3]
# Function Call
ans = constructArray(arr, N)
print(" ".join(list(map(str, ans))))
'''  Code is written by RAJAT KUMAR '''


C#




// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
  static int binary_search(List<int> arr, int low, int high, int x){
    // Check base case
    if (high >= low){
      int mid = (high + low); // 2
      // If element is present at the middle itself
      if (arr[mid] == x){
        return mid;
 
        // If element is smaller than mid, then it can only
        // be present in left subarray
      }else if(arr[mid] > x){
        return binary_search(arr, low, mid - 1, x);
        // Else the element can only be present in right subarray
 
      }else{
        return binary_search(arr, mid + 1, high, x);
      }
    }
    return -1;
  }
 
  static List<int> constructArray(List<int> arr, int n){
    // sorting the arr and assigning to another array
    List<int> arr2 = new List<int>(arr);
    arr2.Sort();
 
    // Initializing prefix_sum array
    int[] prefix_sum = new int[n];
 
    // first element of prefix_sum and given array will be the same
    prefix_sum[0] = arr2[0];
 
    // iterating through 0 to n-1 and
    // calculating prefix_sum
    for (int i = 1 ; i < n ; i++){
      prefix_sum[i] = prefix_sum[i-1] + arr2[i];
    }
 
    for (int i = 0 ; i < n ; i++){
 
      // storing index of each element of given array
      // in our sorted array using binary search
      int element_index = binary_search(arr2, 0, n - 1, arr[i]);
 
      // storing sum of all smaller elements into smaller
      int smaller = prefix_sum[element_index]-arr[i];
 
      // storing sum of all larger element into larger
      int larger = prefix_sum[n-1]-prefix_sum[element_index];
 
      // multiplying smaller and larger and '
      // storing into arr[i]'
      arr[i] = smaller * larger;
    }
 
    return arr;
  }
 
 
  // Driver code
  public static void Main(string[] args){
 
    int N = 4;
    List<int> arr = new List<int>{8, 4, 9, 3};
 
    // Function Call
    List<int> ans = constructArray(arr, N);
    for(int i = 0 ; i < ans.Count ; i++){
      Console.Write(ans[i] + " ");
    }
    Console.WriteLine("");
 
  }
}
 
// This code is contributed entertain2022.


Javascript




// JavaScript code to implement the approach
function binary_search(arr, low, high, x)
{
    // Check base case
    if (high >= low)
    {
        let mid = Math.floor((high + low) / 2);
         
        // If element is present at the middle itself
        if (arr[mid] == x)
            return mid;
             
        // If element is smaller than mid, then it can only
        // be present in left subarray
        else if (arr[mid] > x)
            return binary_search(arr, low, mid - 1, x);
             
        // Else the element can only be present in right subarray
        else
            return binary_search(arr, mid + 1, high, x);
    }
}
 
function constructArray(arr, n)
{
    // sorting the arr and assigning to another array
    let arr2 = [...arr];
    arr2.sort();
     
    // Initializing prefix_sum array
    prefix_sum = new Array(n).fill(0);
     
    // first element of prefix_sum and given array will be the same
    prefix_sum[0] = arr2[0];
     
    // iterating through 0 to n-1 and
    // calculating prefix_sum
    for (var i = 1; i < n; i++)
        prefix_sum[i] = prefix_sum[i-1]+arr2[i];
     
    let element_index, smaller, larger;
    for (var i = 0; i < n; i++)
    {
        // storing index of each element of given array
        // in our sorted array using binary search
        element_index = binary_search(arr2, 0, n-1, arr[i]);
         
        // storing sum of all smaller elements into smaller
        smaller = prefix_sum[element_index]-arr[i];
         
        // storing sum of all larger element into larger
        larger = prefix_sum[n-1]-prefix_sum[element_index]
         
        // multiplying smaller and larger and '
        // storing into arr[i]'
        arr[i] = smaller*larger;
    }
 
    return arr;
}
 
// Driver Code
let N = 4;
let arr = [8, 4, 9, 3];
 
// Function Call
let ans = constructArray(arr, N);
console.log(ans);
 
// This code is contributed by phasing17.


Output

63 51 0 0

Time Complexity: Steps involved into time requirement:

  • Sorting the array takes :O(n*log(n)) time
  • Finding prefix sum takes O(n) time
  • Calling binary_search function each time inside loop running n time : O(n*logn)

Overall Time Complexity: O(n*logn)

Auxiliary Space: O(n)


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