Given N candies and K people. In the first turn, the first person gets 1 candy, the second gets 2 candies, and so on till K people. In the next turn, the first person gets K+1 candies, the second person gets k+2 candies, and so on. If the number of candies is less than the required number of candies at every turn, then the person receives the remaining number of candies.
The task is to find the total number of candies every person has at the end.
Examples:
Input: N = 7, K = 4
Output: 1 2 3 1
At the first turn, the fourth people has to be given 4 candies, but there is
only 1 left, hence he takes one only.Input: N = 10, K = 3
Output: 5 2 3
At the second turn first one receives 4 and then we have no more candies left.
Recommended Practice
A naive approach is to iterate for every turn and distribute candies accordingly till candies are finished.
Time complexity: O(Number of distributions)
A better approach is to perform every turn in O(1) by calculating sum of natural numbers till the last term of series which will be (turns*k) and subtracting the sum of natural numbers till the last term of previous series which is (turns-1)*k. Keep doing this till the sum is less than N, once it exceeds then distribute candies in the given way till possible. We call a turn completed if every person gets the desired number of candies he is to get in a turn.
Below is the implementation of the above approach:
C++
// C++ code for better approach
// to distribute candies
#include <bits/stdc++.h>
using namespace std;
// Function to find out the number of
// candies every person received
void candies(int n, int k)
{
// Count number of complete turns
int count = 0;
// Get the last term
int ind = 1;
// Stores the number of candies
int arr[k];
memset(arr, 0, sizeof(arr));
while (n) {
// Last term of last and
// current series
int f1 = (ind - 1) * k;
int f2 = ind * k;
// Sum of current and last series
int sum1 = (f1 * (f1 + 1)) / 2;
int sum2 = (f2 * (f2 + 1)) / 2;
// Sum of current series only
int res = sum2 - sum1;
// If sum of current is less than N
if (res <= n) {
count++;
n -= res;
ind++;
}
else // Individually distribute
{
int i = 0;
// First term
int term = ((ind - 1) * k) + 1;
// Distribute candies till there
while (n > 0) {
// Candies available
if (term <= n) {
arr[i++] = term;
n -= term;
term++;
}
else // Not available
{
arr[i++] = n;
n = 0;
}
}
}
}
// Count the total candies
for (int i = 0; i < k; i++)
arr[i] += (count * (i + 1))
+ (k * (count * (count - 1)) / 2);
// Print the total candies
for (int i = 0; i < k; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
int n = 10, k = 3;
candies(n, k);
return 0;
}
Java
// Java code for better approach
// to distribute candies
class GFG {
// Function to find out the number of
// candies every person received
static void candies(int n, int k){
int[] arr = new int[k];
int j = 0;
while(n>0){
for(int i =0;i<k;i++){
j++;
if(n<=0){
break;
}
else{
if(j<n){
arr[i] = arr[i]+j;
}
else{
arr[i] = arr[i]+n;
}
n = n-j;
}
}
}
for(int i:arr){
System.out.print(i+" ");
}
}
// Driver Code
public static void main(String[] args)
{
int n = 10, k = 3;
candies(n, k);
}
}
// This code is contributed by ihritik
Python3
# Python3 code for better approach
# to distribute candies
import math as mt
# Function to find out the number of
# candies every person received
def candies(n, k):
# Count number of complete turns
count = 0
# Get the last term
ind = 1
# Stores the number of candies
arr = [0 for i in range(k)]
while n > 0:
# Last term of last and
# current series
f1 = (ind - 1) * k
f2 = ind * k
# Sum of current and last series
sum1 = (f1 * (f1 + 1)) // 2
sum2 = (f2 * (f2 + 1)) //2
# Sum of current series only
res = sum2 - sum1
# If sum of current is less than N
if (res <= n):
count += 1
n -= res
ind += 1
else: # Individually distribute
i = 0
# First term
term = ((ind - 1) * k) + 1
# Distribute candies till there
while (n > 0):
# Candies available
if (term <= n):
arr[i] = term
i += 1
n -= term
term += 1
else:
arr[i] = n
i += 1
n = 0
# Count the total candies
for i in range(k):
arr[i] += ((count * (i + 1)) +
(k * (count * (count - 1)) // 2))
# Print the total candies
for i in range(k):
print(arr[i], end = " ")
# Driver Code
n, k = 10, 3
candies(n, k)
# This code is contributed by Mohit kumar
C#
// C# code for better approach
// to distribute candies
using System;
class GFG
{
// Function to find out the number of
// candies every person received
static void candies(int n, int k)
{
// Count number of complete turns
int count = 0;
// Get the last term
int ind = 1;
// Stores the number of candies
int []arr=new int[k];
for(int i=0;i<k;i++)
arr[i]=0;
while (n>0) {
// Last term of last and
// current series
int f1 = (ind - 1) * k;
int f2 = ind * k;
// Sum of current and last series
int sum1 = (f1 * (f1 + 1)) / 2;
int sum2 = (f2 * (f2 + 1)) / 2;
// Sum of current series only
int res = sum2 - sum1;
// If sum of current is less than N
if (res <= n) {
count++;
n -= res;
ind++;
}
else // Individually distribute
{
int i = 0;
// First term
int term = ((ind - 1) * k) + 1;
// Distribute candies till there
while (n > 0) {
// Candies available
if (term <= n) {
arr[i++] = term;
n -= term;
term++;
}
else // Not available
{
arr[i++] = n;
n = 0;
}
}
}
}
// Count the total candies
for (int i = 0; i < k; i++)
arr[i] += (count * (i + 1))
+ (k * (count * (count - 1)) / 2);
// Print the total candies
for (int i = 0; i < k; i++)
Console.Write( arr[i] + " ");
}
// Driver Code
public static void Main()
{
int n = 10, k = 3;
candies(n, k);
}
}
// This code is contributed by ihritik
PHP
<?php
// PHP code for better approach
// to distribute candies
// Function to find out the number of
// candies every person received
function candies($n, $k)
{
// Count number of complete turns
$count = 0;
// Get the last term
$ind = 1;
// Stores the number of candies
$arr = array_fill(0, $k, 0) ;
while ($n)
{
// Last term of last and
// current series
$f1 = ($ind - 1) * $k;
$f2 = $ind * $k;
// Sum of current and last series
$sum1 = floor(($f1 * ($f1 + 1)) / 2);
$sum2 = floor(($f2 * ($f2 + 1)) / 2);
// Sum of current series only
$res = $sum2 - $sum1;
// If sum of current is less than N
if ($res <= $n)
{
$count++;
$n -= $res;
$ind++;
}
else // Individually distribute
{
$i = 0;
// First term
$term = (($ind - 1) * $k) + 1;
// Distribute candies till there
while ($n > 0)
{
// Candies available
if ($term <= $n)
{
$arr[$i++] = $term;
$n -= $term;
$term++;
}
else // Not available
{
$arr[$i++] = $n;
$n = 0;
}
}
}
}
// Count the total candies
for ($i = 0; $i < $k; $i++)
$arr[$i] += floor(($count * ($i + 1)) + ($k *
($count * ($count - 1)) / 2));
// Print the total candies
for ($i = 0; $i < $k; $i++)
echo $arr[$i], " ";
}
// Driver Code
$n = 10;
$k = 3;
candies($n, $k);
// This code is contributed by Ryuga
?>
Javascript
<script>
// JavaScript code for better approach
// to distribute candies
// Function to find out the number of
// candies every person received
function candies(n , k) {
// Count number of complete turns
var count = 0;
// Get the last term
var ind = 1;
// Stores the number of candies
var arr = Array(k);
for (i = 0; i < k; i++)
arr[i] = 0;
while (n > 0) {
// Last term of last and
// current series
var f1 = (ind - 1) * k;
var f2 = ind * k;
// Sum of current and last series
var sum1 = (f1 * (f1 + 1)) / 2;
var sum2 = (f2 * (f2 + 1)) / 2;
// Sum of current series only
var res = sum2 - sum1;
// If sum of current is less than N
if (res <= n) {
count++;
n -= res;
ind++;
} else // Individually distribute
{
var i = 0;
// First term
var term = ((ind - 1) * k) + 1;
// Distribute candies till there
while (n > 0) {
// Candies available
if (term <= n) {
arr[i++] = term;
n -= term;
term++;
} else // Not available
{
arr[i++] = n;
n = 0;
}
}
}
}
// Count the total candies
for (i = 0; i < k; i++)
arr[i] += (count * (i + 1)) +
(k * (count * (count - 1)) / 2);
// Print the total candies
for (i = 0; i < k; i++)
document.write(arr[i] + " ");
}
// Driver Code
var n = 10, k = 3;
candies(n, k);
// This code contributed by Rajput-Ji
</script>
Output:
5 2 3
Time complexity: O(Number of turns + K)
Auxiliary Space: O(k)
An efficient approach is to find the largest number(say MAXI) whose sum upto natural numbers is less than N using Binary search. Since the last number will always be a multiple of K, we get the last number of complete turns. Subtract the summation till then from N. Distribute the remaining candies by traversing in the array.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find out the number of
// candies every person received
void candies(int n, int k)
{
// Count number of complete turns
int count = 0;
// Get the last term
int ind = 1;
// Stores the number of candies
int arr[k];
memset(arr, 0, sizeof(arr));
int low = 0, high = n;
// Do a binary search to find the number whose
// sum is less than N.
while (low <= high) {
// Get mide
int mid = (low + high) >> 1;
int sum = (mid * (mid + 1)) >> 1;
// If sum is below N
if (sum <= n) {
// Find number of complete turns
count = mid / k;
// Right halve
low = mid + 1;
}
else {
// Left halve
high = mid - 1;
}
}
// Last term of last complete series
int last = (count * k);
// Subtract the sum till
n -= (last * (last + 1)) / 2;
int i = 0;
// First term of incomplete series
int term = (count * k) + 1;
while (n) {
if (term <= n) {
arr[i++] = term;
n -= term;
term++;
}
else {
arr[i] += n;
n = 0;
}
}
// Count the total candies
for (int i = 0; i < k; i++)
arr[i] += (count * (i + 1))
+ (k * (count * (count - 1)) / 2);
// Print the total candies
for (int i = 0; i < k; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
int n = 7, k = 4;
candies(n, k);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to find out the number of
// candies every person received
static void candies(int n, int k)
{
// Count number of complete turns
int count = 0;
// Get the last term
int ind = 1;
// Stores the number of candies
int []arr=new int[k];
for(int i=0;i<k;i++)
arr[i]=0;
int low = 0, high = n;
// Do a binary search to find the number whose
// sum is less than N.
while (low <= high) {
// Get mide
int mid = (low + high) >> 1;
int sum = (mid * (mid + 1)) >> 1;
// If sum is below N
if (sum <= n) {
// Find number of complete turns
count = mid / k;
// Right halve
low = mid + 1;
}
else {
// Left halve
high = mid - 1;
}
}
// Last term of last complete series
int last = (count * k);
// Subtract the sum till
n -= (last * (last + 1)) / 2;
int j = 0;
// First term of incomplete series
int term = (count * k) + 1;
while (n > 0) {
if (term <= n) {
arr[j++] = term;
n -= term;
term++;
}
else {
arr[j] += n;
n = 0;
}
}
// Count the total candies
for (int i = 0; i < k; i++)
arr[i] += (count * (i + 1))
+ (k * (count * (count - 1)) / 2);
// Print the total candies
for (int i = 0; i < k; i++)
System.out.print( arr[i] + " " );
}
// Driver Code
public static void main(String []args)
{
int n = 7, k = 4;
candies(n, k);
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of the above approach
# Function to find out the number of
# candies every person received
def candies(n, k):
# Count number of complete turns
count = 0;
# Get the last term
ind = 1;
# Stores the number of candies
arr = [0] * k;
low = 0;
high = n;
# Do a binary search to find the
# number whose sum is less than N.
while (low <= high):
# Get mide
mid = (low + high) >> 1;
sum = (mid * (mid + 1)) >> 1;
# If sum is below N
if (sum <= n):
# Find number of complete turns
count = int(mid / k);
# Right halve
low = mid + 1;
else:
# Left halve
high = mid - 1;
# Last term of last complete series
last = (count * k);
# Subtract the sum till
n -= int((last * (last + 1)) / 2);
i = 0;
# First term of incomplete series
term = (count * k) + 1;
while (n):
if (term <= n):
arr[i] = term;
i += 1;
n -= term;
term += 1;
else:
arr[i] += n;
n = 0;
# Count the total candies
for i in range(k):
arr[i] += ((count * (i + 1)) +
int(k * (count * (count - 1)) / 2));
# Print the total candies
for i in range(k):
print(arr[i], end = " ");
# Driver Code
n = 7;
k = 4;
candies(n, k);
# This code is contributed by chandan_jnu
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to find out the number of
// candies every person received
static void candies(int n, int k)
{
// Count number of complete turns
int count = 0;
// Get the last term
int ind = 1;
// Stores the number of candies
int []arr=new int[k];
for(int i=0;i<k;i++)
arr[i]=0;
int low = 0, high = n;
// Do a binary search to find the number whose
// sum is less than N.
while (low <= high) {
// Get mide
int mid = (low + high) >> 1;
int sum = (mid * (mid + 1)) >> 1;
// If sum is below N
if (sum <= n) {
// Find number of complete turns
count = mid / k;
// Right halve
low = mid + 1;
}
else {
// Left halve
high = mid - 1;
}
}
// Last term of last complete series
int last = (count * k);
// Subtract the sum till
n -= (last * (last + 1)) / 2;
int j = 0;
// First term of incomplete series
int term = (count * k) + 1;
while (n > 0) {
if (term <= n) {
arr[j++] = term;
n -= term;
term++;
}
else {
arr[j] += n;
n = 0;
}
}
// Count the total candies
for (int i = 0; i < k; i++)
arr[i] += (count * (i + 1))
+ (k * (count * (count - 1)) / 2);
// Print the total candies
for (int i = 0; i < k; i++)
Console.Write( arr[i] + " " );
}
// Driver Code
public static void Main()
{
int n = 7, k = 4;
candies(n, k);
}
}
// This code is contributed by ihritik
PHP
<?php
// PHP implementation of the above approach
// Function to find out the number of
// candies every person received
function candies($n, $k)
{
// Count number of complete turns
$count = 0;
// Get the last term
$ind = 1;
// Stores the number of candies
$arr = array_fill(0, $k, 0);
$low = 0;
$high = $n;
// Do a binary search to find the
// number whose sum is less than N.
while ($low <= $high)
{
// Get mide
$mid = ($low + $high) >> 1;
$sum = ($mid * ($mid + 1)) >> 1;
// If sum is below N
if ($sum <= $n)
{
// Find number of complete turns
$count = (int)($mid / $k);
// Right halve
$low = $mid + 1;
}
else
{
// Left halve
$high = $mid - 1;
}
}
// Last term of last complete series
$last = ($count * $k);
// Subtract the sum till
$n -= (int)(($last * ($last + 1)) / 2);
$i = 0;
// First term of incomplete series
$term = ($count * $k) + 1;
while ($n)
{
if ($term <= $n)
{
$arr[$i++] = $term;
$n -= $term;
$term++;
}
else
{
$arr[$i] += $n;
$n = 0;
}
}
// Count the total candies
for ($i = 0; $i < $k; $i++)
$arr[$i] += ($count * ($i + 1)) +
(int)($k * ($count * ($count - 1)) / 2);
// Print the total candies
for ($i = 0; $i < $k; $i++)
echo $arr[$i] . " ";
}
// Driver Code
$n = 7;
$k = 4;
candies($n, $k);
// This code is contributed
// by chandan_jnu
?>
Javascript
<script>
// javascript implementation of the above approach
// Function to find out the number of
// candies every person received
function candies(n , k) {
// Count number of complete turns
var count = 0;
// Get the last term
var ind = 1;
// Stores the number of candies
var arr = Array(k).fill(0);
for (i = 0; i < k; i++)
arr[i] = 0;
var low = 0, high = n;
// Do a binary search to find the number whose
// sum is less than N.
while (low <= high) {
// Get mide
var mid = parseInt((low + high) /2);
var sum = parseInt((mid * (mid + 1)) / 2);
// If sum is below N
if (sum <= n) {
// Find number of complete turns
count = parseInt(mid / k);
// Right halve
low = mid + 1;
} else {
// Left halve
high = mid - 1;
}
}
// Last term of last complete series
var last = (count * k);
// Subtract the sum till
n -= (last * (last + 1)) / 2;
var j = 0;
// First term of incomplete series
var term = (count * k) + 1;
while (n > 0) {
if (term <= n) {
arr[j++] = term;
n -= term;
term++;
} else {
arr[j] += n;
n = 0;
}
}
// Count the total candies
for (i = 0; i < k; i++)
arr[i] += (count * (i + 1)) + (k * (count * (count - 1)) / 2);
// Print the total candies
for (i = 0; i < k; i++)
document.write(arr[i] + " ");
}
// Driver Code
var n = 7, k = 4;
candies(n, k);
// This code contributed by aashish1995
</script>
Output:
1 2 3 1
Time Complexity: O(log N + K)
Auxiliary Space: O(K) for given K
Distribute N candies among K people in c++:
Approach:
The distribute_candies function takes two integers as input: N, which represents the total number of candies, and K, which represents the number of people. It returns a vector of integers, where the i-th element represents the number of candies distributed to the i-th person.
The function initializes a vector result of K elements with zero candies. It then loops until all N candies have been distributed. In each iteration, it calculates the number of candies to give to the current person (candies_to_give) as the minimum of N and i+1. It then adds candies_to_give to the number of candies distributed to the i-th person in result, and subtracts candies_to_give from N. Finally, it increments i to move to the next person.
C++
#include <iostream>
#include <vector>
std::vector<int> distribute_candies(int N, int K) {
std::vector<int> result(K, 0); // initialize a vector of K elements with zero candies
int i = 0;
while (N > 0) { // loop until we have no more candies to distribute
int candies_to_give = std::min(N, i+1);
result[i % K] += candies_to_give; // distribute candies to the i-th person
N -= candies_to_give; // subtract the distributed candies from N
i += 1; // move to the next person
}
return result;
}
int main() {
int N = 10;
int K = 3;
std::vector<int> result = distribute_candies(N, K);
for (int i = 0; i < K; i++) {
std::cout << result[i] << " ";
}
return 0;
}
Java
import java.util.*;
public class DistributeCandies {
public static List<Integer> distributeCandies(int N,
int K)
{
List<Integer> result
= new ArrayList<>(Collections.nCopies(
K, 0)); // initialize a list of K elements
// with zero candies
int i = 0;
while (N > 0) { // loop until we have no more
// candies to distribute
int candiesToGive = Math.min(N, i + 1);
result.set(
i % K,
result.get(i % K)
+ candiesToGive); // distribute candies
// to the i-th person
N -= candiesToGive; // subtract the distributed
// candies from N
i += 1; // move to the next person
}
return result;
}
public static void main(String[] args)
{
int N = 10;
int K = 3;
List<Integer> result = distributeCandies(N, K);
for (int i = 0; i < K; i++) {
System.out.print(result.get(i) + " ");
}
}
}
Python3
def distribute_candies(N, K):
result = [0] * K # initialize a list of K elements with zero candies
i = 0
while N > 0: # loop until we have no more candies to distribute
candies_to_give = min(N, i+1)
result[i % K] += candies_to_give # distribute candies to the i-th person
N -= candies_to_give # subtract the distributed candies from N
i += 1 # move to the next person
return result
if __name__ == '__main__':
N = 10
K = 3
result = distribute_candies(N, K)
for i in range(K):
print(result[i], end=" ")
# output: 3 3 4
C#
using System;
using System.Collections.Generic;
public class Gfg {
public static List<int> distribute_candies(int N, int K) {
List<int> result = new List<int>(new int[K]); // initialize a list of K elements with zero candies
int i = 0;
while (N > 0) { // loop until we have no more candies to distribute
int candies_to_give = Math.Min(N, i+1);
result[i % K] += candies_to_give; // distribute candies to the i-th person
N -= candies_to_give; // subtract the distributed candies from N
i += 1; // move to the next person
}
return result;
}
public static void Main() {
int N = 10;
int K = 3;
List<int> result = distribute_candies(N, K);
for (int i = 0; i < K; i++) {
Console.Write(result[i] + " ");
}
// output: 3 3 4
}
}
Javascript
// JavaScript equivalent
function distribute_candies(N, K) {
let result = Array(K).fill(0); // initialize a list of K elements with zero candies
let i = 0;
while (N > 0) { // loop until we have no more candies to distribute
let candies_to_give = Math.min(N, i+1);
result[i % K] += candies_to_give; // distribute candies to the i-th person
N -= candies_to_give; // subtract the distributed candies from N
i += 1; // move to the next person
}
return result;
}
let N = 10;
let K = 3;
let result = distribute_candies(N, K); temp="";
for (let i = 0; i < K; i++) {
temp = temp+result[i]+" ";
} console.log(temp);
Output
5 2 3
The time complexity of this algorithm is O(N), because we need to distribute each of the N candies.
The auxiliary space of this algorithm is O(K), because we use a vector of K elements to store the candies distributed to each person.
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