Replace every element in a circular array by sum of next K elements
Last Updated :
15 Jul, 2025
Given a circular array arr[] of N integers and an integer K, the task is to print the array after the following operations:
- If K is non-negative, then replace A[i] with the sum of the next K elements.
- If K is negative, then replace it with the sum of previous K elements.
A cyclic array means the next element of the last element of the array is the first element and the previous element of the first element is the last element.
Examples:
Input: arr[] = {4, 2, -5, 11}, K = 3
Output: 8 10 17 1
Explanation:
Step 1: For index 0, replace arr[0] = arr[1] + arr[2] + arr[3] = 2 + (-5) + 11 = 8
Step 2: For index 1, replace arr[1] = arr[2] + arr[3] + arr[0] = (-5) + 11 + 4 = 10
Step 3: For index 2, replace arr[2] = arr[3] + arr[0] + arr[1] = 11 + 4 + 2 = 17
Step 4: For index 3, replace arr[3] = arr[0] + arr[1] + arr[2] = 4 + 2 + -5 = 1
Therefore, the resultant array is {8, 10, 17, 1}
Input: arr[] = {2, 5, 7, 9}, K = -2
Output: 16 11 7 12
Explanation:
Step 1: For index 0, replace arr[0] = arr[3] + arr[2] = 9 + 7 = 16
Step 2: For index 1, replace arr[1] = arr[0] + arr[3] = 2 + 9 = 11
Step 3: For index 2, replace arr[2] = arr[1] + arr[0] = 5 + 2 = 7
Step 4: For index 3, replace arr[3] = arr[2] + arr[1] = 7 + 5 = 12
Therefore, the resultant array is {16, 11, 7, 12}
Naive Approach: The simplest approach to solve the problem is to traverse the array and traverse the next or previous K elements, based on the value of K, for every array element and print their sum. Follow the steps below to solve the problem:
- If the value of K is negative, then reverse the array and find the sum of the next K elements.
- If the value of K is non-negative, then simply find the sum of the next K elements for each element.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
// Function to find the sum of next
// or previous k array elements
vector<int> sumOfKelementsUtil(
vector<int>& a, int x)
{
// Size of the array
int n = a.size();
int count, k, temp;
// Stores the result
vector<int> ans;
// Iterate the given array
for (int i = 0; i < n; i++) {
count = 0;
k = i + 1;
temp = 0;
// Traverse the k elements
while (count < x) {
temp += a[k % n];
count++;
k++;
}
// Push the K elements sum
ans.pb(temp);
}
// Return the resultant vector
return ans;
}
// Function that prints the sum of next
// K element for each index in the array
void sumOfKelements(vector<int>& arr,
int K)
{
// Size of the array
int N = (int)arr.size();
// Stores the result
vector<int> ans;
// If key is negative,
// reverse the array
if (K < 0) {
K = K * (-1);
// Reverse the array
reverse(arr.begin(),
arr.end());
// Find the resultant vector
ans = sumOfKelementsUtil(arr, K);
// Reverse the array again to
// get the original sequence
reverse(ans.begin(), ans.end());
}
// If K is positive
else {
ans = sumOfKelementsUtil(arr, K);
}
// Print the answer
for (int i = 0; i < N; i++) {
cout << ans[i] << " ";
}
}
// Driver Code
int main()
{
// Given array arr[]
vector<int> arr = { 4, 2, -5, 11 };
int K = 3;
// Function Call
sumOfKelements(arr, K);
return 0;
}
// Java program for
// the above approach
import java.util.*;
class GFG{
//reverse array
static Vector<Integer> reverse(Vector<Integer> a)
{
int i, n = a.size(), t;
for (i = 0; i < n / 2; i++)
{
t = a.elementAt(i);
a.set(i, a.elementAt(n - i - 1));
a.set(n - i - 1, t);
}
return a;
}
// Function to find the sum of next
// or previous k array elements
static Vector<Integer> sumOfKelementsUtil(
Vector<Integer>a, int x)
{
// Size of the array
int n = a.size();
int count, k, temp;
// Stores the result
Vector<Integer> ans = new Vector<>();
// Iterate the given array
for (int i = 0; i < n; i++)
{
count = 0;
k = i + 1;
temp = 0;
// Traverse the k elements
while (count < x)
{
temp += a.elementAt(k % n);
count++;
k++;
}
// Push the K elements sum
ans.add(temp);
}
// Return the resultant vector
return ans;
}
// Function that prints the sum of next
// K element for each index in the array
static void sumOfKelements(Vector<Integer> arr,
int K)
{
// Size of the array
int N = (int)arr.size();
// Stores the result
Vector<Integer> ans = new Vector<>();
// If key is negative,
// reverse the array
if (K < 0)
{
K = K * (-1);
// Reverse the array
arr = reverse(arr);
// Find the resultant vector
ans = sumOfKelementsUtil(arr, K);
// Reverse the array again to
// get the original sequence
ans = reverse(ans);
}
// If K is positive
else
{
ans = sumOfKelementsUtil(arr, K);
}
// Print the answer
for (int i = 0; i < N; i++)
{
System.out.print(ans.elementAt(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
Vector<Integer> arr = new Vector<>();
arr.add(4);
arr.add(2);
arr.add(-5);
arr.add(11);
int K = 3;
// Function Call
sumOfKelements(arr, K);
}
}
// This code is contributed by gauravrajput1
# Python3 program for
# the above approach
# Function to find the sum of next
# or previous k array elements
def sumOfKelementsUtil(a, x):
# Size of the array
n = len(a)
# Stores the result
ans = []
# Iterate the given array
for i in range(n):
count = 0
k = i + 1
temp = 0
# Traverse the k elements
while (count < x):
temp += a[k % n]
count += 1
k += 1
# Push the K elements sum
ans.append(temp)
# Return the resultant vector
return ans
# Function that prints the
# sum of next K element for
# each index in the array
def sumOfKelements(arr, K):
# Size of the array
N = len(arr)
#Stores the result
ans = []
# If key is negative,
# reverse the array
if (K < 0):
K = K * (-1)
# Reverse the array
arr.reverse()
# Find the resultant vector
ans = sumOfKelementsUtil(arr, K)
#Reverse the array again to
# get the original sequence
ans.reverse()
# If K is positive
else:
ans = sumOfKelementsUtil(arr, K)
# Print the answer
for i in range(N):
print (ans[i], end = " ")
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [4, 2, -5, 11]
K = 3
# Function Call
sumOfKelements(arr, K)
# This code is contributed by Chitranayal
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Reverse array
static List<int> reverse(List<int> a)
{
int i, n = a.Count, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// Function to find the sum of next
// or previous k array elements
static List<int> sumOfKelementsUtil(
List<int>a, int x)
{
// Size of the array
int n = a.Count;
int count, k, temp;
// Stores the result
List<int> ans = new List<int>();
// Iterate the given array
for (int i = 0; i < n; i++)
{
count = 0;
k = i + 1;
temp = 0;
// Traverse the k elements
while (count < x)
{
temp += a[k % n];
count++;
k++;
}
// Push the K elements sum
ans.Add(temp);
}
// Return the resultant vector
return ans;
}
// Function that prints the sum of next
// K element for each index in the array
static void sumOfKelements(List<int> arr,
int K)
{
// Size of the array
int N = (int)arr.Count;
// Stores the result
List<int> ans = new List<int>();
// If key is negative,
// reverse the array
if (K < 0)
{
K = K * (-1);
// Reverse the array
arr = reverse(arr);
// Find the resultant vector
ans = sumOfKelementsUtil(arr, K);
// Reverse the array again to
// get the original sequence
ans = reverse(ans);
}
// If K is positive
else
{
ans = sumOfKelementsUtil(arr, K);
}
// Print the answer
for (int i = 0; i < N; i++)
{
Console.Write(ans[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
List<int> arr = new List<int>();
arr.Add(4);
arr.Add(2);
arr.Add(-5);
arr.Add(11);
int K = 3;
// Function Call
sumOfKelements(arr, K);
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript program for the above approach
// Function to print the
// required resultant array
function sumOfKElements(
arr, n, k)
{
// Reverse the array
let rev = false;
if (k < 0) {
rev = true;
k *= -1;
let l = 0, r = n - 1;
// Traverse the range
while (l < r) {
let tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l++;
r--;
}
}
// Store prefix sum
let dp = Array.from({length: n}, (_, i) => 0);
dp[0] = arr[0];
// Find the prefix sum
for (let i = 1; i < n; i++) {
dp[i] += dp[i - 1] + arr[i];
}
// Store the answer
let ans = Array.from({length: n}, (_, i) => 0);
// Calculate the answers
for (let i = 0; i < n; i++) {
if (i + k < n)
ans[i] = dp[i + k] - dp[i];
else {
// Count of remaining elements
let x = k - (n - 1 - i);
// Add the sum of all elements
// y times
let y = Math.floor(x / n);
// Add the remaining elements
let rem = x % n;
// Update ans[i]
ans[i] = dp[n - 1]
- dp[i]
+ y * dp[n - 1]
+ (rem - 1 >= 0 ? dp[rem - 1] : 0);
}
}
// If array is reversed print
// ans[] in reverse
if (rev) {
for (let i = n - 1; i >= 0; i--) {
document.write(ans[i] + " ");
}
}
else {
for (let i = 0; i < n; i++) {
document.write(ans[i] + " ");
}
}
}
// Driver Code
// Given array arr[]
let arr = [ 4, 2, -5, 11 ];
let N = arr.length;
// Given K
let K = 3;
// Function
sumOfKElements(arr, N, K);
// This code is contributed by souravghosh0416.
</script>
Time Complexity: O(N*K), where N is the length of the given array and K is the given integer.
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use prefix sum. Follow the steps below to solve the problem:
- If K is negative, reverse the given array and multiply K with -1.
- Compute and store the prefix sum of the given array in pre[].
- Create the array ans[] to store the answer for each element.
- For every index i, if i + K is smaller than N, then update ans[i] = pre[i + k] - pre[i]
- Otherwise, add the sum of all the elements from index i to N - 1, find remaining K - (N - 1 - i) elements because (N - 1 - i) elements are already added in the above step.
- Add the sum of all elements floor((K - N - 1 - i)/N) times and the sum of remaining elements of the given array from 0 to ((K - (N - 1 - i)) % N) - 1.
- After calculating the answer for each element of the array, check if the array was reversed previously. If yes, reverse the ans[] array.
- Print all the elements stored in the array ans[] after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to print the
// required resultant array
void sumOfKElements(int arr[], int n,
int k)
{
// Reverse the array
bool rev = false;
if (k < 0)
{
rev = true;
k *= -1;
int l = 0, r = n - 1;
// Traverse the range
while (l < r)
{
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l++;
r--;
}
}
// Store prefix sum
int dp[n] = {0};
dp[0] = arr[0];
// Find the prefix sum
for(int i = 1; i < n; i++)
{
dp[i] += dp[i - 1] + arr[i];
}
// Store the answer
int ans[n] = {0};
// Calculate the answers
for(int i = 0; i < n; i++)
{
if (i + k < n)
ans[i] = dp[i + k] - dp[i];
else
{
// Count of remaining elements
int x = k - (n - 1 - i);
// Add the sum of all elements
// y times
int y = x / n;
// Add the remaining elements
int rem = x % n;
// Update ans[i]
ans[i] = dp[n - 1] - dp[i] +
y * dp[n - 1] + (rem - 1 >= 0 ?
dp[rem - 1] : 0);
}
}
// If array is reversed print
// ans[] in reverse
if (rev)
{
for(int i = n - 1; i >= 0; i--)
{
cout << ans[i] << " ";
}
}
else
{
for(int i = 0; i < n; i++)
{
cout << ans[i] << " ";
}
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 4, 2, -5, 11 };
int N = sizeof(arr) / sizeof(arr[0]);
// Given K
int K = 3;
// Function
sumOfKElements(arr, N, K);
}
// This code is contributed by SURENDRA_GANGWAR
// Java program for the above approach
import java.io.*;
class GFG {
// Function to print the
// required resultant array
static void sumOfKElements(
int arr[], int n, int k)
{
// Reverse the array
boolean rev = false;
if (k < 0) {
rev = true;
k *= -1;
int l = 0, r = n - 1;
// Traverse the range
while (l < r) {
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l++;
r--;
}
}
// Store prefix sum
int dp[] = new int[n];
dp[0] = arr[0];
// Find the prefix sum
for (int i = 1; i < n; i++) {
dp[i] += dp[i - 1] + arr[i];
}
// Store the answer
int ans[] = new int[n];
// Calculate the answers
for (int i = 0; i < n; i++) {
if (i + k < n)
ans[i] = dp[i + k] - dp[i];
else {
// Count of remaining elements
int x = k - (n - 1 - i);
// Add the sum of all elements
// y times
int y = x / n;
// Add the remaining elements
int rem = x % n;
// Update ans[i]
ans[i] = dp[n - 1]
- dp[i]
+ y * dp[n - 1]
+ (rem - 1 >= 0 ? dp[rem - 1] : 0);
}
}
// If array is reversed print
// ans[] in reverse
if (rev) {
for (int i = n - 1; i >= 0; i--) {
System.out.print(ans[i] + " ");
}
}
else {
for (int i = 0; i < n; i++) {
System.out.print(ans[i] + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 4, 2, -5, 11 };
int N = arr.length;
// Given K
int K = 3;
// Function
sumOfKElements(arr, N, K);
}
}
# Python3 program for
# the above approach
# Function to print
# required resultant array
def sumOfKElements(arr, n, k):
# Reverse the array
rev = False;
if (k < 0):
rev = True;
k *= -1;
l = 0;
r = n - 1;
# Traverse the range
while (l < r):
tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l += 1;
r -= 1;
# Store prefix sum
dp = [0] * n;
dp[0] = arr[0];
# Find the prefix sum
for i in range(1, n):
dp[i] += dp[i - 1] + arr[i];
# Store the answer
ans = [0] * n;
# Calculate the answers
for i in range(n):
if (i + k < n):
ans[i] = dp[i + k] - dp[i];
else:
# Count of remaining
# elements
x = k - (n - 1 - i);
# Add the sum of
# all elements y times
y = x // n;
# Add the remaining
# elements
rem = x % n;
# Update ans[i]
ans[i] = (dp[n - 1] - dp[i] +
y * dp[n - 1] +
(dp[rem - 1]
if rem - 1 >= 0
else 0));
# If array is reversed
# print ans in reverse
if (rev):
for i in range(n - 1, -1, -1):
print(ans[i], end = " ");
else:
for i in range(n):
print(ans[i], end = " ");
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [4, 2, -5, 11];
N = len(arr);
# Given K
K = 3;
# Function
sumOfKElements(arr, N, K);
# This code is contributed by 29AjayKumar
// C# program for
// the above approach
using System;
class GFG {
// Function to print the
// required resultant array
static void sumOfKElements(int []arr,
int n, int k)
{
// Reverse the array
bool rev = false;
if (k < 0)
{
rev = true;
k *= -1;
int l = 0, r = n - 1;
// Traverse the range
while (l < r)
{
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l++;
r--;
}
}
// Store prefix sum
int []dp = new int[n];
dp[0] = arr[0];
// Find the prefix sum
for (int i = 1; i < n; i++)
{
dp[i] += dp[i - 1] + arr[i];
}
// Store the answer
int []ans = new int[n];
// Calculate the answers
for (int i = 0; i < n; i++)
{
if (i + k < n)
ans[i] = dp[i + k] - dp[i];
else
{
// Count of remaining elements
int x = k - (n - 1 - i);
// Add the sum of all elements
// y times
int y = x / n;
// Add the remaining elements
int rem = x % n;
// Update ans[i]
ans[i] = dp[n - 1] - dp[i] +
y * dp[n - 1] +
(rem - 1 >= 0 ?
dp[rem - 1] : 0);
}
}
// If array is reversed print
// ans[] in reverse
if (rev)
{
for (int i = n - 1; i >= 0; i--)
{
Console.Write(ans[i] + " ");
}
}
else
{
for (int i = 0; i < n; i++)
{
Console.Write(ans[i] + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = {4, 2, -5, 11};
int N = arr.Length;
// Given K
int K = 3;
// Function
sumOfKElements(arr, N, K);
}
}
// This code is contributed by Princi Singh
<script>
// Javascript program for the above approach
// Function to print the
// required resultant array
function sumOfKElements(arr, n,
k)
{
// Reverse the array
let rev = false;
if (k < 0)
{
rev = true;
k *= -1;
let l = 0, r = n - 1;
// Traverse the range
while (l < r)
{
let tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
l++;
r--;
}
}
// Store prefix sum
let dp = new Uint8Array(n);
dp[0] = arr[0];
// Find the prefix sum
for(let i = 1; i < n; i++)
{
dp[i] += dp[i - 1] + arr[i];
}
// Store the answer
let ans = new Uint8Array(n);
// Calculate the answers
for(let i = 0; i < n; i++)
{
if (i + k < n)
ans[i] = dp[i + k] - dp[i];
else
{
// Count of remaining elements
let x = k - (n - 1 - i);
// Add the sum of all elements
// y times
let y = Math.floor(x / n);
// Add the remaining elements
let rem = x % n;
// Update ans[i]
ans[i] = dp[n - 1] - dp[i] +
y * dp[n - 1] + (rem - 1 >= 0 ?
dp[rem - 1] : 0);
}
}
// If array is reversed print
// ans[] in reverse
if (rev)
{
for(let i = n - 1; i >= 0; i--)
{
document.write(ans[i] + " ");
}
}
else
{
for(let i = 0; i < n; i++)
{
document.write(ans[i] + " ");
}
}
}
// Driver Code
// Given array arr[]
let arr = [ 4, 2, -5, 11 ];
let N = arr.length;
// Given K
let K = 3;
// Function
sumOfKElements(arr, N, K);
//This code is contributed by Mayank Tyagi
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem