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.
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 receivedvoid 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 Codeint main(){ int n = 10, k = 3; candies(n, k); return 0;} |
Java
// Java code for better approach// to distribute candiesclass 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 candiesimport math as mt# Function to find out the number of# candies every person receiveddef 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 Coden, k = 10, 3candies(n, k)# This code is contributed by Mohit kumar |
C#
// C# code for better approach// to distribute candiesusing 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 receivedvoid 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 Codeint main(){ int n = 7, k = 4; candies(n, k); return 0;} |
Java
// Java implementation of the above approachclass 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 receiveddef 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 Coden = 7;k = 4;candies(n, k);# This code is contributed by chandan_jnu |
C#
// C# implementation of the above approachusing 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 receivedfunction 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 resultif __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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