Sum of elements in range L-R where first half and second half is filled with odd and even numbers
Given a number N, create an array such the first half of the array is filled with odd numbers till N and second half of the array is filled with even numbers. Also given are L and R indices, the task is to print the sum of elements in the array in the range [L, R].
Examples:
Input: N = 12, L = 1, R = 11
Output: 66
The array formed thus is {1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10, 12}
The sum between index 1 and index 11 is 66 {1+3+5+7+9+11+2+4+6+8+10}Input: N = 11, L = 3 and R = 7
Output: 34
The array formed is {1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10}
The sum between index 3 and index 7 is 34 {5+7+9+11+2}
A naive approach will be to form the array of size N and iterate from L to R and return the sum.
C++
// C++ program to find the sum between L-R // range by creating the array // Naive Approach #include<bits/stdc++.h> using namespace std; // Function to find the sum between L and R int rangesum(int n, int l, int r) { // array created int arr[n]; // fill the first half of array int c = 1, i = 0; while (c <= n) { arr[i++] = c; c += 2; } // fill the second half of array c = 2; while (c <= n) { arr[i++] = c; c += 2; } int sum = 0; // find the sum between range for (i = l - 1; i < r; i++) { sum += arr[i]; } return sum; } // Driver Code int main() { int n = 12; int l = 1, r = 11; cout<<(rangesum(n, l, r)); } // This code is contributed by // Sanjit_Prasad |
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Java
// Java program to find the sum between L-R // range by creating the array // Naive Approach import java.io.*; public class GFG { // Function to find the sum between L and R static int rangesum(int n, int l, int r) { // array created int[] arr = new int[n]; // fill the first half of array int c = 1, i = 0; while (c <= n) { arr[i++] = c; c += 2; } // fill the second half of array c = 2; while (c <= n) { arr[i++] = c; c += 2; } int sum = 0; // find the sum between range for (i = l - 1; i < r; i++) { sum += arr[i]; } return sum; } // Driver Code public static void main(String[] args) { int n = 12; int l = 1, r = 11; System.out.println(rangesum(n, l, r)); } } |
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Python3
# Python3 program to find the sum between L-R # range by creating the array # Naive Approach # Function to find the sum between L and R def rangesum(n, l, r): # array created arr = [0] * n; # fill the first half of array c = 1; i = 0; while (c <= n): arr[i] = c; i += 1; c += 2; # fill the second half of array c = 2; while (c <= n): arr[i] = c; i += 1; c += 2; sum = 0; # find the sum between range for i in range(l - 1, r, 1): sum += arr[i]; return sum; # Driver Code if __name__ == '__main__': n = 12; l, r = 1, 11; print(rangesum(n, l, r)); # This code contributed by PrinciRaj1992 |
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C#
// C# program to find the sum // between L-R range by creating // the array Naive Approach using System; class GFG { // Function to find the // sum between L and R static int rangesum(int n, int l, int r) { // array created int[] arr = new int[n]; // fill the first // half of array int c = 1, i = 0; while (c <= n) { arr[i++] = c; c += 2; } // fill the second // half of array c = 2; while (c <= n) { arr[i++] = c; c += 2; } int sum = 0; // find the sum // between range for (i = l - 1; i < r; i++) { sum += arr[i]; } return sum; } // Driver Code public static void Main() { int n = 12; int l = 1, r = 11; Console.WriteLine(rangesum(n, l, r)); } } // This code is contributed // by inder_verma. |
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PHP
<?php // PHP program to find the sum between // L-R range by creating the array // Naive Approach // Function to find the sum between L and R function rangesum($n, $l, $r) { // array created $arr = array_fill(0, $n, 0); // fill the first half of array $c = 1; $i = 0; while ($c <= $n) { $arr[$i++] = $c; $c += 2; } // fill the second half of array $c = 2; while ($c <= $n) { $arr[$i++] = $c; $c += 2; } $sum = 0; // find the sum between range for ($i = $l - 1; $i < $r; $i++) { $sum += $arr[$i]; } return $sum; } // Driver Code $n = 12; $l = 1; $r = 11; echo(rangesum($n, $l, $r)); // This code is contributed by mits ?> |
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Output:
66
Time complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach: Without building the array the problem can be solved in O(1) complexity. Consider the middle point of the sequence thus formed. Then there can be 3 possibilities for L and R.
- l and r both are on the right of middle point.
- l and r both are on the left of middle point.
- l is on the left of the middle point and r is on the left of the middle point.
For 1st and 2nd case, just find the number which corresponds to the l-th position from the start or middle and the number which corresponds to the r-th position from start or middle. The below formula can be used to find out the sum:
sum = (no. of terms in the range)*(first term + last term)/2
For the third case, consider it as [L-mid] and [mid-R] and then apply case 1 and case 2 formula as mentioned above.
Below is the implementation of the above approach:
C++
// C++ program to find the sum between L-R // range by creating the array // Naive Approach #include<bits/stdc++.h> using namespace std; // // Function to calculate the sum if n is even int sumeven(int n, int l, int r) { int sum = 0; int mid = n / 2; // both l and r are to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // Total number of terms in // the sequence is r-l+1 int no_of_terms = r - l + 1; // use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // both l and r are to the right of mid else if (l >= mid) { // first and last element int first = (2 * (l - n / 2)); int last = (2 * (r - n / 2)); int no_of_terms = r - l + 1; // Use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // left is to the left of mid and // right is to the right of mid else { // Take two sums i.e left and // right differently and add int sumleft = 0, sumright = 0; // first and last element int first_term1 = (2 * l - 1); int last_term1 = (2 * (n / 2) - 1); // total terms int no_of_terms1 = n / 2 - l + 1; // no of terms sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // The first even number is 2 int first_term2 = 2; // The last element is given by 2*(r-n/2) int last_term2 = (2 * (r - n / 2)); int no_of_terms2 = r - mid; // formula applied sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; sum = (sumleft + sumright); } return sum; } // Function to calculate the sum if n is odd int sumodd(int n, int l, int r) { // take ceil value if n is odd int mid = n / 2 + 1; int sum = 0; // // both l and r are to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // number of terms int no_of_terms = r - l + 1; // formula sum = ((no_of_terms) * ((first + last))) / 2; } // // both l and r are to the right of mid else if (l > mid) { // firts and last ter, int first = (2 * (l - mid)); int last = (2 * (r - mid)); // no of terms int no_of_terms = r - l + 1; // formula used sum = ((no_of_terms) * ((first + last))) / 2; } // If l is on left and r on right else { // calculate separate sums int sumleft = 0, sumright = 0; // first half int first_term1 = (2 * l - 1); int last_term1 = (2 * mid - 1); // calculate terms int no_of_terms1 = mid - l + 1; sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // second half int first_term2 = 2; int last_term2 = (2 * (r - mid)); int no_of_terms2 = r - mid; sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; // add both halves sum = (sumleft + sumright); } return sum; } // Function to find the sum between L and R int rangesum(int n, int l, int r) { int sum = 0; // If n is even if (n % 2 == 0) return sumeven(n, l, r); // If n is odd else return sumodd(n, l, r); } // Driver Code int main() { int n = 12; int l = 1, r = 11; cout << (rangesum(n, l, r)); } // This code is contributed by 29AjayKumar |
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Java
// Java program to find the sum between L-R // range by creating the array // Efficient Approach import java.io.*; public class GFG { // // Function to calculate the sum if n is even static int sumeven(int n, int l, int r) { int sum = 0; int mid = n / 2; // both l and r are to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // Total number of terms in // the sequence is r-l+1 int no_of_terms = r - l + 1; // use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // both l and r are to the right of mid else if (l >= mid) { // // first and last element int first = (2 * (l - n / 2)); int last = (2 * (r - n / 2)); int no_of_terms = r - l + 1; // Use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // left is to the left of mid and // right is to the right of mid else { // Take two sums i.e left and // right differently and add int sumleft = 0, sumright = 0; // first and last element int first_term1 = (2 * l - 1); int last_term1 = (2 * (n / 2) - 1); // total terms int no_of_terms1 = n / 2 - l + 1; // no of terms sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // The first even number is 2 int first_term2 = 2; // The last element is given by 2*(r-n/2) int last_term2 = (2 * (r - n / 2)); int no_of_terms2 = r - mid; // formula applied sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; sum = (sumleft + sumright); } return sum; } // Function to calculate the sum if n is odd static int sumodd(int n, int l, int r) { // take ceil value if n is odd int mid = n / 2 + 1; int sum = 0; // // both l and r are to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // number of terms int no_of_terms = r - l + 1; // formula sum = ((no_of_terms) * ((first + last))) / 2; } // // both l and r are to the right of mid else if (l > mid) { // firts and last ter, int first = (2 * (l - mid)); int last = (2 * (r - mid)); // no of terms int no_of_terms = r - l + 1; // formula used sum = ((no_of_terms) * ((first + last))) / 2; } // If l is on left and r on right else { // calculate separate sums int sumleft = 0, sumright = 0; // first half int first_term1 = (2 * l - 1); int last_term1 = (2 * mid - 1); // calculate terms int no_of_terms1 = mid - l + 1; sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // second half int first_term2 = 2; int last_term2 = (2 * (r - mid)); int no_of_terms2 = r - mid; sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; // add both halves sum = (sumleft + sumright); } return sum; } // Function to find the sum between L and R static int rangesum(int n, int l, int r) { int sum = 0; // If n is even if (n % 2 == 0) return sumeven(n, l, r); // If n is odd else return sumodd(n, l, r); } // Driver Code public static void main(String[] args) { int n = 12; int l = 1, r = 11; System.out.println(rangesum(n, l, r)); } } |
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C#
// C# program to find the sum // between L-R range by creating // the array Efficient Approach using System; class GFG { // Function to calculate // the sum if n is even static int sumeven(int n, int l, int r) { int sum = 0; int mid = n / 2; // both l and r are // to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // Total number of terms in // the sequence is r-l+1 int no_of_terms = r - l + 1; // use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // both l and r are // to the right of mid else if (l >= mid) { // first and last element int first = (2 * (l - n / 2)); int last = (2 * (r - n / 2)); int no_of_terms = r - l + 1; // Use of formula derived sum = ((no_of_terms) * ((first + last))) / 2; } // left is to the left of // mid and right is to the // right of mid else { // Take two sums i.e left and // right differently and add int sumleft = 0, sumright = 0; // first and last element int first_term1 = (2 * l - 1); int last_term1 = (2 * (n / 2) - 1); // total terms int no_of_terms1 = n / 2 - l + 1; // no of terms sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // The first even // number is 2 int first_term2 = 2; // The last element is // given by 2*(r-n/2) int last_term2 = (2 * (r - n / 2)); int no_of_terms2 = r - mid; // formula applied sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; sum = (sumleft + sumright); } return sum; } // Function to calculate // the sum if n is odd static int sumodd(int n, int l, int r) { // take ceil value // if n is odd int mid = n / 2 + 1; int sum = 0; // both l and r are // to the left of mid if (r <= mid) { // first and last element int first = (2 * l - 1); int last = (2 * r - 1); // number of terms int no_of_terms = r - l + 1; // formula sum = ((no_of_terms) * ((first + last))) / 2; } // both l and r are // to the right of mid else if (l > mid) { // firts and last ter, int first = (2 * (l - mid)); int last = (2 * (r - mid)); // no of terms int no_of_terms = r - l + 1; // formula used sum = ((no_of_terms) * ((first + last))) / 2; } // If l is on left // and r on right else { // calculate separate sums int sumleft = 0, sumright = 0; // first half int first_term1 = (2 * l - 1); int last_term1 = (2 * mid - 1); // calculate terms int no_of_terms1 = mid - l + 1; sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2; // second half int first_term2 = 2; int last_term2 = (2 * (r - mid)); int no_of_terms2 = r - mid; sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2; // add both halves sum = (sumleft + sumright); } return sum; } // Function to find the // sum between L and R static int rangesum(int n, int l, int r) { // If n is even if (n % 2 == 0) return sumeven(n, l, r); // If n is odd else return sumodd(n, l, r); } // Driver Code public static void Main() { int n = 12; int l = 1, r = 11; Console.WriteLine(rangesum(n, l, r)); } } // This code is contributed // by chandan_jnu. |
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Output:
66
Time complexity: O(1)
Auxiliary Space: O(1)
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