Replace every element in a circular array by sum of next K elements
Last Updated :
09 Sep, 2021
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
DSA Tutorial - Learn Data Structures and Algorithms
DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on
7 min read
Quick Sort
QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s
12 min read
Merge Sort - Data Structure and Algorithms Tutorials
Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge
14 min read
SQL Commands | DDL, DQL, DML, DCL and TCL Commands
SQL commands are crucial for managing databases effectively. These commands are divided into categories such as Data Definition Language (DDL), Data Manipulation Language (DML), Data Control Language (DCL), Data Query Language (DQL), and Transaction Control Language (TCL). In this article, we will e
7 min read
Breadth First Search or BFS for a Graph
Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
15+ min read
Bubble Sort Algorithm
Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir
8 min read
Data Structures Tutorial
Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st
2 min read
Binary Search Algorithm - Iterative and Recursive Implementation
Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
Insertion Sort Algorithm
Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T
9 min read
Dijkstra's Algorithm to find Shortest Paths from a Source to all
Given a weighted undirected graph represented as an edge list and a source vertex src, find the shortest path distances from the source vertex to all other vertices in the graph. The graph contains V vertices, numbered from 0 to V - 1.Note: The given graph does not contain any negative edge. Example
12 min read