Sum of all odd length subarrays

18 Aug, 2021
Given an array arr[] consisting of N integers, the task is to find the sum of all the elements of all possible subarrays of odd length.
Examples:
Input: arr[] = {3, 2, 4}Output: 18Explanation:The odd length subarrays along with their sum are as follows:1) {3} = sum is 3.2) {2} = sum is 2.3) {4} = sum is 4.4) {3, 2, 4} = sum is 3 + 2 + 4 = 9.Therefore, sum of all subarrays = 3 + 2 + 4 + 9 = 18.
Input: arr[] = {1, 2, 1, 2}Output: 15Explanation:The odd length subarrays along with their sum are as follows:1) {1} = sum is 1.2) {2} = sum is 2.3) {1} = sum is 1.4) {2} = sum is 2.5) {1, 2, 1} = sum is 1 + 2 + 1 = 4.6) {2, 1, 2} = sum is 2 + 1 + 2 = 5.Therefore, sum of all subarrays = 1 + 2 + 1 + 2 + 4 + 5 = 15.
Naive Approach: The simplest approach is to generate all possible subarrays of odd length from the given array and find the sum of all such subarrays.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Function to calculate the sum// of all odd length subarraysint OddLengthSum(vector<int>& arr){    // Stores the sum    int sum = 0;     // Size of array    int l = arr.size();     // Traverse the array    for (int i = 0; i < l; i++) {         // Generate all subarray of        // odd length        for (int j = i; j < l; j += 2) {             for (int k = i; k <= j; k++) {                 // Add the element to sum                sum += arr[k];            }        }    }     // Return the final sum    return sum;} // Driver Codeint main(){    // Given array arr[]    vector<int> arr = { 1, 5, 3, 1, 2 };     // Function Call    cout << OddLengthSum(arr);     return 0;}
// Java program for the above approachimport java.io.*;import java.util.Arrays; class GFG{  // Function to calculate the sum// of all odd length subarraysstatic int OddLengthSum(int[] arr){         // Stores the sum    int sum = 0;      // Size of array    int l = arr.length;      // Traverse the array    for(int i = 0; i < l; i++)    {                 // Generate all subarray of        // odd length        for(int j = i; j < l; j += 2)        {            for(int k = i; k <= j; k++)            {                                 // Add the element to sum                sum += arr[k];            }        }    }         // Return the final sum    return sum;}  // Driver Codepublic static void main (String[] args){         // Given array arr[]    int[] arr = { 1, 5, 3, 1, 2 };      // Function call    System.out.print(OddLengthSum(arr));}} // This code is contributed by sanjoy_62
# Python3 program for the above approach  # Function to calculate the sum# of all odd length subarraysdef OddLengthSum(arr):         # Stores the sum    sum = 0      # Size of array    l = len(arr)      # Traverse the array    for i in range(l):          # Generate all subarray of        # odd length        for j in range(i, l, 2):            for k in range(i, j + 1, 1):                  # Add the element to sum                sum += arr[k]                 # Return the final sum    return sum # Driver Code # Given array arr[]arr = [ 1, 5, 3, 1, 2 ]  # Function callprint(OddLengthSum(arr)) # This code is contributed by sanjoy_62
// C# program for the above approachusing System; class GFG{  // Function to calculate the sum// of all odd length subarraysstatic int OddLengthSum(int[] arr){         // Stores the sum    int sum = 0;      // Size of array    int l = arr.Length;      // Traverse the array    for(int i = 0; i < l; i++)    {                 // Generate all subarray of        // odd length        for(int j = i; j < l; j += 2)        {            for(int k = i; k <= j; k++)            {                                 // Add the element to sum                sum += arr[k];            }        }    }      // Return the final sum    return sum;}  // Driver Codepublic static void Main (){         // Given array arr[]    int[] arr = { 1, 5, 3, 1, 2 };      // Function call    Console.Write(OddLengthSum(arr));}} // This code is contributed by sanjoy_62
<script>// javascript program for the above approach  // Function to calculate the sum// of all odd length subarraysfunction OddLengthSum(arr){          // Stores the sum    var sum = 0;       // Size of array    var l = arr.length;       // Traverse the array    for(var i = 0; i < l; i++)    {                  // Generate all subarray of        // odd length        for(var j = i; j < l; j += 2)        {            for(var k = i; k <= j; k++)            {                                  // Add the element to sum                sum += arr[k];            }        }    }       // Return the final sum    return sum;}   // Driver Code     // Given array arr[]    var arr = [ 1, 5, 3, 1, 2 ]       // Function call    document.write(OddLengthSum(arr)); // This code is contributed by bunnyram19.</script>
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Time Complexity: O(N3) Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to observe the following pattern after generating all the subarrays of odd length:
For any element at index idx there are (idx + 1) choices on the left side of it and (N – idx) choices on the right side of it.
Therefore, for any element arr[i], the count of arr[i] is (i + 1) * (N – i) in all the subarrays.
So, for an element arr[i], there are ((i + 1) * (N – i) + 1) / 2 sub-arrays with odd length.
Finally, arr[i] will have a total of ((i + 1) * (n – i) + 1) / 2 frequency in the sum.
Therefore, to find the sum of all elements of all the subarrays of odd length, the idea is to iterate over the array and for every ith array element, add [((i + 1) * (n – i) + 1) / 2]*arr[i] to the sum.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function that finds the sum of all// the element of subarrays of odd lengthint OddLengthSum(vector<int>& arr){    // Stores the sum    int sum = 0;     // Size of array    int l = arr.size();     // Traverse the given array arr[]    for (int i = 0; i < l; i++) {         // Add to the sum for each        // contribution of the arr[i]        sum += (((i + 1)                     * (l - i)                 + 1)                / 2)               * arr[i];    }     // Return the final sum    return sum;} // Driver Codeint main(){    // Given array arr[]    vector<int> arr = { 1, 5, 3, 1, 2 };     // Function Call    cout << OddLengthSum(arr);     return 0;}
// Java program for the above approachclass GFG{ // Function that finds the sum of all// the element of subarrays of odd lengthstatic int OddLengthSum(int []arr){         // Stores the sum    int sum = 0;     // Size of array    int l = arr.length;     // Traverse the given array arr[]    for(int i = 0; i < l; i++)    {                 // Add to the sum for each        // contribution of the arr[i]        sum += (((i + 1) * (l - i) +                 1) / 2) * arr[i];    }     // Return the final sum    return sum;} // Driver Codepublic static void main(String[] args){         // Given array arr[]    int []arr = { 1, 5, 3, 1, 2 };     // Function call    System.out.print(OddLengthSum(arr));}} // This code is contributed by Amit Katiyar
# Python3 program for the above approach # Function that finds the sum of all# the element of subarrays of odd lengthdef OddLengthSum(arr):     # Stores the sum    Sum = 0     # Size of array    l = len(arr)     # Traverse the given array arr[]    for i in range(l):             # Add to the sum for each        # contribution of the arr[i]        Sum += ((((i + 1) *                  (l - i) +                 1) // 2) * arr[i])         # Return the final sum    return Sum # Driver code # Given array arr[]arr = [ 1, 5, 3, 1, 2 ] # Function callprint(OddLengthSum(arr)) # This code is contributed by divyeshrabadiya07
// C# program for// the above approachusing System;class GFG{ // Function that finds the// sum of all the element of// subarrays of odd lengthstatic int OddLengthSum(int []arr){  // Stores the sum  int sum = 0;   // Size of array  int l = arr.Length;   // Traverse the given array []arr  for(int i = 0; i < l; i++)  {    // Add to the sum for each    // contribution of the arr[i]    sum += (((i + 1) *             (l - i) + 1) / 2) * arr[i];  }   // Return the readonly sum  return sum;} // Driver Codepublic static void Main(String[] args){  // Given array []arr  int []arr = {1, 5, 3, 1, 2};   // Function call  Console.Write(OddLengthSum(arr));}} // This code is contributed by Rajput-Ji
<script> // Javascript program to implement// the above approach // Function that finds the sum of all// the element of subarrays of odd lengthfunction OddLengthSum(arr){          // Stores the sum    let sum = 0;      // Size of array    let l = arr.length;      // Traverse the given array arr[]    for(let i = 0; i < l; i++)    {                  // Add to the sum for each        // contribution of the arr[i]        sum += Math.floor(((i + 1) * (l - i) +                 1) / 2) * arr[i];    }      // Return the final sum    return sum;} // Driver Code     // Given array arr[]    let arr = [ 1, 5, 3, 1, 2 ];      // Function call    document.write(OddLengthSum(arr)); // This code is contributed by target_2.</script>
48
Time Complexity: O(N) Auxiliary Space: O(1)
sanjoy_62
amit143katiyar
Rajput-Ji
divyeshrabadiya07
bunnyram19
target_2
binod kumar singh
subarray
subarray-sum
Arrays
Mathematical
Pattern Searching
Arrays
Mathematical
Pattern Searching
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