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Construct sum-array with sum of elements in given range

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  • Last Updated : 02 Aug, 2022
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You are given an array of n-elements and an odd-integer m. You have to construct a new sum_array from given array such that sum_array[i] = Σarr[j] for (i-(m/2)) < j (i+(m/2))

note : for 0 > j or j >= n take arr[j] = 0.

Examples: 

Input : arr[] = {1, 2, 3, 4, 5}, 
            m = 3
Output : sum_array = {3, 6, 9, 12, 9}
Explanation : sum_array[0] = arr[0] + arr[1]
     sum_array[1] = arr[0] + arr[1] + arr[2]
     sum_array[2] = arr[1] + arr[2] + arr[3]
     sum_array[3] = arr[2] + arr[3] + arr[4]
     sum_array[4] = arr[3] + arr[4]

Input : arr[] = {2, 4, 3, 4, 2}, 
           m = 1
Output : sum_array = {2, 4, 3, 4, 2}
Explanation : sum_array[0] = arr[0] 
              sum_array[1] = arr[1]
              sum_array[2] = arr[2]
              sum_array[3] = arr[3]
              sum_array[4] = arr[4]

Basic Approach : As per problem statement, we calculate sum_array[i] by iterating over i-(m/2) to i+(m/2). According to this approach, we have a nested loop which will result into time complexity of O(n*m).
Efficient Approach : For calculating sum_array is to use sliding window concept and thus can easily save our time. For Sliding window, the time complexity is O(n). 

Algorithm 

calculate sum of first (m/2)+1 elementssum_array[0] = sumfor i=1 to i<nif( (i-(m/2)-1) >= 0 )
           sum -= arr[(i-(m/2)-1)]if( (i+m/2) < n)
           sum += arr[(i+m/2)]sum_array[i] = sumprint sum_array

Implementation:

C++




// CPP Program to find sum array for a given
// array.
#include <bits/stdc++.h>
using namespace std;
 
// function to calc sum_array and print
void calcSum_array(int arr[], int n, int m)
{
    int sum = 0;
    int sum_array[n];
 
    // calc 1st m/2 + 1 element for 1st window
    for (int i = 0; i < m / 2 + 1; i++)
        sum += arr[i];
    sum_array[0] = sum;
 
    // use sliding window to
    // calculate rest of sum_array
    for (int i = 1; i < n; i++) {
        if (i - (m / 2) - 1 >= 0)
            sum -= arr[i - (m / 2) - 1];
        if (i + (m / 2) < n)
            sum += arr[i + (m / 2)];
        sum_array[i] = sum;
    }
 
    // print sum_array
    for (int i = 0; i < n; i++)
        cout << sum_array[i] << " ";
}
 
// driver program
int main()
{
    int arr[] = { 3, 6, 2, 7, 3, 8, 4,
                      9, 1, 5, 0, 4 };
    int m = 5;
    int n = sizeof(arr) / sizeof(int);
    calcSum_array(arr, n, m);
    return 0;
}

Java




// Java Program to find sum array
// for a given array.
class GFG
{
    // function to calc sum_array and print
    static void calcSum_array(int arr[], int n, int m)
    {
        int sum = 0;
        int sum_array[] = new int[n];
     
        // calc 1st m/2 + 1 element
        // for 1st window
        for (int i = 0; i < m / 2 + 1; i++)
            sum += arr[i];
        sum_array[0] = sum;
     
        // use sliding window to
        // calculate rest of sum_array
        for (int i = 1; i < n; i++)
        {
            if (i - (m / 2) - 1 >= 0)
                sum -= arr[i - (m / 2) - 1];
            if (i + (m / 2) < n)
                sum += arr[i + (m / 2)];
            sum_array[i] = sum;
        }
     
        // print sum_array
        for (int i = 0; i < n; i++)
            System.out.print(sum_array[i] + " ");
    }
     
    // Driver program
    public static void main(String[] args)
    {
        int arr[] = { 3, 6, 2, 7, 3, 8, 4, 9, 1, 5, 0, 4 };
        int m = 5;
        int n = arr.length;
        calcSum_array(arr, n, m);
    }
}
 
// This code is contributed by prerna saini.

Python3




# Python3 Program to find Sum array
# for a given array.
import math as mt
 
# function to calc Sum_array and print
def calcSum_array(arr, n, m):
 
    Sum = 0
    Sum_array = [0 for i in range(n)]
 
    # calc 1st m/2 + 1 element for 1st window
    for i in range(m // 2 + 1):
        Sum += arr[i]
    Sum_array[0] = Sum
 
    # use sliding window to
    # calculate rest of Sum_array
    for i in range(1, n):
        if (i - (m // 2) - 1 >= 0):
            Sum -= arr[i - (m // 2) - 1]
        if (i + (m / 2) < n):
            Sum += arr[i + (m //2)]
        Sum_array[i] = Sum
     
    # print Sum_array
    for i in range(n):
        print(Sum_array[i], end = " ")
 
# Driver Code
arr = [ 3, 6, 2, 7, 3, 8, 4, 9, 1, 5, 0, 4 ]
m = 5
n = len(arr)
calcSum_array(arr, n, m)
 
# This code is contributed by mohit kumar 29

C#




// C# Program to find sum array
// for a given array.
using System;
 
class GFG
{
    // function to calc sum_array and print
    static void calcSum_array(int []arr, int n, int m)
    {
        int sum = 0;
        int []sum_array = new int[n];
     
        // calc 1st m/2 + 1 element
        // for 1st window
        for (int i = 0; i < m / 2 + 1; i++)
            sum += arr[i];
        sum_array[0] = sum;
     
        // use sliding window to
        // calculate rest of sum_array
        for (int i = 1; i < n; i++)
        {
            if (i - (m / 2) - 1 >= 0)
                sum -= arr[i - (m / 2) - 1];
            if (i + (m / 2) < n)
                sum += arr[i + (m / 2)];
            sum_array[i] = sum;
        }
     
        // print sum_array
        for (int i = 0; i < n; i++)
        Console.Write(sum_array[i] + " ");
    }
     
    // Driver program
    public static void Main()
    {
        int []arr = { 3, 6, 2, 7, 3, 8, 4, 9, 1, 5, 0, 4 };
        int m = 5;
        int n = arr.Length;
        calcSum_array(arr, n, m);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Program to find sum array
// for a given array.
 
// function to calc sum_array and print
function calcSum_array(&$arr, $n, $m)
{
    $sum = 0;
    $sum_array = array();
 
    // calc 1st m/2 + 1 element
    // for 1st window
    for ($i = 0;
         $i < (int)($m / 2) + 1; $i++)
        $sum = $sum + $arr[$i];
    $sum_array[0] = $sum;
 
    // use sliding window to
    // calculate rest of sum_array
    for ($i = 1; $i < $n; $i++)
    {
        if ($i - (int)($m / 2) - 1 >= 0)
            $sum = $sum - $arr[$i -
                    (int)($m / 2) - 1];
        if ($i + (int)($m / 2) < $n)
            $sum = $sum + $arr[$i +
                    (int)($m / 2)];
        $sum_array[$i] = $sum;
    }
 
    // print sum_array
    for ($i = 0; $i < $n; $i++)
        echo $sum_array[$i] . " ";
}
 
// Driver Code
$arr = array(3, 6, 2, 7, 3, 8,
             4, 9, 1, 5, 0, 4 );
$m = 5;
$n = sizeof($arr);
calcSum_array($arr, $n, $m);
 
// This code is contributed by Mukul Singh
?>

Javascript




<script>
// JavaScript Program to find sum array for a given
// array.
 
// function to calc sum_array and print
function calcSum_array(arr, n, m)
{
    let sum = 0;
    let sum_array = new Array(n);
 
    // calc 1st m/2 + 1 element for 1st window
    for (let i = 0; i < Math.floor(m / 2) + 1; i++)
        sum += arr[i];
    sum_array[0] = sum;
 
    // use sliding window to
    // calculate rest of sum_array
    for (let i = 1; i < n; i++)
    {
        if (i - Math.floor(m / 2) - 1 >= 0)
            sum -= arr[i - Math.floor(m / 2) - 1];
        if (i + Math.floor(m / 2) < n)
            sum += arr[i + Math.floor(m / 2)];
        sum_array[i] = sum;
    }
 
    // print sum_array
    for (let i = 0; i < n; i++)
        document.write(sum_array[i] + " ");
}
 
// Driver program
 
    let arr = [ 3, 6, 2, 7, 3, 8, 4,
                    9, 1, 5, 0, 4 ];
    let m = 5;
    let n = arr.length;
    calcSum_array(arr, n, m);
 
// This code is contributed by Surbhi Tyagi.
</script>

Output

11 18 21 26 24 31 25 27 19 19 10 9 

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