Given two integer arrays ages and packs where ages store the ages of different students and an element of pack stores the number of candies that packet has (complete array represent the number of packets). The candies can be distributed among students such that:
- Every student must get only one pack of candies.
- All students of the same age must get equal number of candies.
- A student which is older must get more candies than all the student who are younger than him.
The task is to determine whether it is possible to distribute candies in the described manner. If possible then print Yes else print No.
Examples:
Input: ages[] = {5, 15, 10}, packs[] = {2, 2, 2, 3, 3, 4}
Output: YES
There are 3 students with age 5, 15 and 10.And there are 6 packets of candies containing 2, 2, 2, 3, 3, 4 candies respectively.
We will give one packet containing 2 candies to the student of age 5, one packet containing 3 candies to student with age 10 and give the packet containing 4 candies to student age 15Input: ages[] = {5, 5, 6, 7}, packs[] = {5, 4, 6, 6}
Output: NO
Approach:
- Make 2 frequency arrays, one which will store the number of students with a particular age and one which will store the number of packets with a particular amount of candies.
- Then traverse the frequency array for ages starting from the youngest age and for every age in ascending try to find the candy packets that are greater than or equal to the number of students for the selected age (starting from the least number of candies a packet has)
- If the above case fails then the answer is No else repeat the above steps until all the student have got the candies and print Yes in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h>using namespace std;// Function to check The validity of distribution void check_distribution(int n, int k, int age[], int candy[]){ // Storing the max age of all // students + 1 int mxage = *(std::max_element( age, age + n)) + 1; // Storing the max candy + 1 int mxcandy = *(std::max_element( candy, candy + k)) + 1; int fr1[mxage] = {0}; int fr2[mxcandy] = {0}; // Creating the frequency array of // the age of students for(int j = 0; j < n; j++) { fr1[age[j]] += 1; } // Creating the frequency array of the // packets of candies for(int j = 0; j < k; j++) { fr2[candy[j]] += 1; } // Pointer to tell whether we have reached // the end of candy frequency array k = 0; // Flag to tell if distribution // is possible or not bool Tf = true; for(int j = 0; j < mxage; j++) { if (fr1[j] == 0) continue; // Flag to tell if we can choose // some candy packets for the // students with age j bool flag = false; while (k < mxcandy) { // If the quantity of packets is // greater than or equal to the // number of students of age j, // then we can choose these // packets for the students if (fr1[j] <= fr2[k]) { flag = true; break; } k += 1; } // Start searching from k + 1 // in next operation k = k + 1; // If we cannot choose any packets // then the answer is NO if (flag == false) { Tf = false; break; } } if (Tf) cout << "YES" << endl; else cout << "NO" << endl; }// Driver code int main(){ int age[] = { 5, 15, 10 }; int candy[] = { 2, 2, 2, 3, 3, 4 }; int n = sizeof(age) / sizeof(age[0]); int k = sizeof(candy) / sizeof(candy[0]); check_distribution(n, k, age, candy); return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java implementation of the approachimport java.util.*;class Main{ // Function to check The validity of distribution public static void check_distribution(int n, int k, int age[], int candy[]) { // Storing the max age of all // students + 1 int mxage = age[0]; for(int i = 0; i < age.length; i++) { if(mxage < age[i]) { mxage = age[i]; } } // Storing the max candy + 1 int mxcandy = candy[0]; for(int i = 0; i < candy.length; i++) { if(mxcandy < candy[i]) { mxcandy = candy[i]; } } int fr1[] = new int[mxage + 1]; Arrays.fill(fr1, 0); int fr2[] = new int[mxcandy + 1]; Arrays.fill(fr2, 0); // Creating the frequency array of // the age of students for(int j = 0; j < n; j++) { fr1[age[j]] += 1; } // Creating the frequency array of the // packets of candies for(int j = 0; j < k; j++) { fr2[candy[j]] += 1; } // Pointer to tell whether we have reached // the end of candy frequency array k = 0; // Flag to tell if distribution // is possible or not boolean Tf = true; for(int j = 0; j < mxage; j++) { if (fr1[j] == 0) continue; // Flag to tell if we can choose // some candy packets for the // students with age j boolean flag = false; while (k < mxcandy) { // If the quantity of packets is // greater than or equal to the // number of students of age j, // then we can choose these // packets for the students if (fr1[j] <= fr2[k]) { flag = true; break; } k += 1; } // Start searching from k + 1 // in next operation k = k + 1; // If we cannot choose any packets // then the answer is NO if (flag == false) { Tf = false; break; } } if (Tf) System.out.println("YES"); else System.out.println("NO"); } // Driver code public static void main(String[] args) { int age[] = { 5, 15, 10 }; int candy[] = { 2, 2, 2, 3, 3, 4 }; int n = age.length; int k = candy.length; check_distribution(n, k, age, candy); }}// This code is contributed by divyesh072019 |
Python3
# Python3 implementation of the approach# Function to check The validity of distributiondef check_distribution(n, k, age, candy): # Storing the max age of all students + 1 mxage = max(age)+1 # Storing the max candy + 1 mxcandy = max(candy)+1 fr1 = [0] * mxage fr2 = [0] * mxcandy # creating the frequency array of the # age of students for j in range(n): fr1[age[j]] += 1 # Creating the frequency array of the # packets of candies for j in range(k): fr2[candy[j]] += 1 # pointer to tell whether we have reached # the end of candy frequency array k = 0 # Flag to tell if distribution is possible or not Tf = True for j in range(mxage): if (fr1[j] == 0): continue # Flag to tell if we can choose some # candy packets for the students with age j flag = False while (k < mxcandy): # If the quantity of packets is greater # than or equal to the number of students # of age j, then we can choose these # packets for the students if (fr1[j] <= fr2[k]): flag = True break k += 1 # Start searching from k + 1 in next operation k = k + 1 # If we cannot choose any packets # then the answer is NO if (flag == False): Tf = False break if (Tf): print("YES") else: print("NO")# Driver codeage = [5, 15, 10]candy = [2, 2, 2, 3, 3, 4]n = len(age)k = len(candy)check_distribution(n, k, age, candy) |
C#
// C# implementation of the approachusing System.IO;using System;class GFG{ // Function to check The validity of distribution static void check_distribution(int n,int k, int[] age,int[] candy) { // Storing the max age of all // students + 1 int mxage = age[0]; for(int i = 0; i < age.Length; i++) { if(mxage < age[i]) { mxage = age[i]; } } // Storing the max candy + 1 int mxcandy = candy[0]; for(int i = 0; i < candy.Length; i++) { if(mxcandy < candy[i]) { mxcandy = candy[i]; } } int[] fr1 = new int[mxage + 1]; Array.Fill(fr1, 0); int[] fr2 = new int[mxcandy + 1]; Array.Fill(fr2, 0); // Creating the frequency array of // the age of students for(int j = 0; j < n; j++) { fr1[age[j]] += 1; } // Creating the frequency array of the // packets of candies for(int j = 0; j < k; j++) { fr2[candy[j]] += 1; } // Pointer to tell whether we have reached // the end of candy frequency array k = 0; // Flag to tell if distribution // is possible or not bool Tf = true; for(int j = 0; j < mxage; j++) { if(fr1[j] == 0) { continue; } // Flag to tell if we can choose // some candy packets for the // students with age j bool flag = false; while (k < mxcandy) { // If the quantity of packets is // greater than or equal to the // number of students of age j, // then we can choose these // packets for the students if (fr1[j] <= fr2[k]) { flag = true; break; } k += 1; } // Start searching from k + 1 // in next operation k = k + 1; // If we cannot choose any packets // then the answer is NO if (flag == false) { Tf = false; break; } } if(Tf) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } // Driver code static void Main() { int[] age = {5, 15, 10}; int[] candy = { 2, 2, 2, 3, 3, 4 }; int n = age.Length; int k = candy.Length; check_distribution(n, k, age, candy); }}// This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // Javascript implementation of the approach // Function to check The validity of distribution function check_distribution(n, k, age, candy) { // Storing the max age of all // students + 1 let mxage = age[0]; for(let i = 0; i < age.length; i++) { if(mxage < age[i]) { mxage = age[i]; } } // Storing the max candy + 1 let mxcandy = candy[0]; for(let i = 0; i < candy.length; i++) { if(mxcandy < candy[i]) { mxcandy = candy[i]; } } let fr1 = new Array(mxage + 1); fr1.fill(0); let fr2 = new Array(mxcandy + 1); fr2.fill(0); // Creating the frequency array of // the age of students for(let j = 0; j < n; j++) { fr1[age[j]] += 1; } // Creating the frequency array of the // packets of candies for(let j = 0; j < k; j++) { fr2[candy[j]] += 1; } // Pointer to tell whether we have reached // the end of candy frequency array k = 0; // Flag to tell if distribution // is possible or not let Tf = true; for(let j = 0; j < mxage; j++) { if(fr1[j] == 0) { continue; } // Flag to tell if we can choose // some candy packets for the // students with age j let flag = false; while (k < mxcandy) { // If the quantity of packets is // greater than or equal to the // number of students of age j, // then we can choose these // packets for the students if (fr1[j] <= fr2[k]) { flag = true; break; } k += 1; } // Start searching from k + 1 // in next operation k = k + 1; // If we cannot choose any packets // then the answer is NO if (flag == false) { Tf = false; break; } } if(Tf) { document.write("YES"); } else { document.write("NO"); } } let age = [5, 15, 10]; let candy = [ 2, 2, 2, 3, 3, 4 ]; let n = age.length; let k = candy.length; check_distribution(n, k, age, candy); // This code is contributed by suresh07.</script> |
Output:
YES
Time Complexity: O(maximum(n, k, max(age)))
Space Complexity: O(max(age))
