Skip to content
Related Articles

Related Articles

Distribution of a Number in Array within a Range
  • Last Updated : 07 Aug, 2020

Given integers S, N, K, L and R where S has to be distributed in an array of size N such that each element must be from the range [L, R] and the sum of K elements of the array should be greater than the sum of the remaining N – K elements whose sum is equal to Sk and these elements are in non-increasing order.

Examples:

Input: N = 5, K = 3, L = 1, R = 3, S = 13, Sk = 9
Output: 3 3 3 2 2

Input: N = 5, K = 3, L = 1, R = 3, S = 15, Sk = 9
Output: 3 3 3 3 3

Approach: If Sk can be distributed into k elements equally then store Sk/k into all the first k elements of the array, otherwise first element of the array will be (Sk/k) + (Sk % k) and remaining k – 1 elements will be (Sk – Sk % k) % k – 1. Similarly, distribute the S-Sk into n-k elements.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function for the distribution of the number
void distribution(int n, int k, int l,
                  int r, int S, int Sk)
{
    int a[n];
    int len = k, temp, rem, s;
    int diff = S - Sk;
    for (int i = 0; i < len; i++) {
        // Distribute the number
        // among k elements
        temp = Sk / k;
        rem = Sk % k;
        if (temp + rem >= l && temp + rem <= r) {
            a[i] = temp;
        }
        else if (temp + rem > r) {
            a[i] = r;
        }
        else if (temp + rem < r) {
            cout << "-1";
            return;
        }
        Sk = Sk - a[i];
        k = k - 1;
    }
  
    // If there is some remaining
    // sum to distribute
    if (Sk > 0) {
        cout << "-1";
        return;
    }
  
    // If there are elements remaining
    // to distribute i.e. (n - k)
    if (len) {
        k = n - len;
        for (int i = len; i < n; i++) {
            // Divide the remaining sum into
            // n-k elements
            temp = diff / k;
            rem = diff % k;
            if (temp + rem >= l && temp + rem <= r) {
                a[i] = temp;
            }
            else if (temp + rem > r) {
                a[i] = r;
            }
            else if (temp + rem < r) {
                cout << "-1";
                return;
            }
            diff = diff - a[i];
            k = k - 1;
        }
        if (diff) {
            cout << "-1";
            return;
        }
    }
  
    // Print the distribution
    for (int i = 0; i < n; i++) {
        cout << a[i] << " ";
    }
}
  
// Driver code
int main()
{
    int n = 5, k = 3, l = 1,
        r = 5, S = 13, Sk = 9;
  
    distribution(n, k, l, r, S, Sk);
  
    return 0;
}

Java




// Java implementation of the approach 
class GFG 
{
      
    // Function for the distribution of the number 
    static void distribution(int n, int k, int l, 
                            int r, int S, int Sk) 
    
        int a[] = new int[n]; 
        int len = k, temp, rem, s; 
        int diff = S - Sk; 
          
        for (int i = 0; i < len; i++)
        {
              
            // Distribute the number 
            // among k elements 
            temp = Sk / k; 
            rem = Sk % k; 
            if (temp + rem >= l && temp + rem <= r)
            
                a[i] = temp; 
            
            else if (temp + rem > r)
            
                a[i] = r; 
            
            else if (temp + rem < r)
            
                System.out.print("-1"); 
                return
            
            Sk = Sk - a[i]; 
            k = k - 1
        
      
        // If there is some remaining 
        // sum to distribute 
        if (Sk > 0)
        
            System.out.print("-1"); 
            return
        
      
        // If there are elements remaining 
        // to distribute i.e. (n - k) 
        if (len != 0)
        
            k = n - len; 
            for (int i = len; i < n; i++) 
            {
                  
                // Divide the remaining sum into 
                // n-k elements 
                temp = diff / k; 
                rem = diff % k; 
                if (temp + rem >= l && temp + rem <= r)
                
                    a[i] = temp; 
                
                else if (temp + rem > r)
                
                    a[i] = r; 
                
                else if (temp + rem < r) 
                
                    System.out.print("-1"); 
                    return
                
                diff = diff - a[i]; 
                k = k - 1
            
            if (diff != 0
            
                System.out.print("-1"); 
                return
            
        
      
        // Print the distribution 
        for (int i = 0; i < n; i++)
        
            System.out.print(a[i] + " "); 
        
    
      
    // Driver code 
    public static void main (String[] args)
    
        int n = 5, k = 3, l = 1
            r = 5, S = 13, Sk = 9
      
        distribution(n, k, l, r, S, Sk); 
    
}
  
// This code is contributed by AnkitRai01

Python3




# Python implementation of the approach 
  
# Function for the distribution of the number
def distribution(n, k, l, r, S, Sk):
    a = [0] * n;
    len = k;
    temp, rem, s = 0, 0, 0;
    diff = S - Sk;
  
    for i in range(len):
  
        # Distribute the number
        # among k elements
        temp = Sk / k;
        rem = Sk % k;
        if (temp + rem >= l and temp + rem <= r):
            a[i] = temp;
        elif(temp + rem > r):
            a[i] = r;
        elif(temp + rem < r):
            print("-1");
            return;
          
        Sk = Sk - a[i];
        k = k - 1;
      
    # If there is some remaining
    # sum to distribute
    if (Sk > 0):
        print("-1");
        return;
      
    # If there are elements remaining
    # to distribute i.e. (n - k)
    if (len != 0):
        k = n - len;
        for i in range(len, n):
  
            # Divide the remaining sum into
            # n-k elements
            temp = diff / k;
            rem = diff % k;
            if (temp + rem >= l and temp + rem <= r):
                a[i] = temp;
            elif(temp + rem > r):
                a[i] = r;
            elif(temp + rem < r):
                print("-1");
                return;
              
            diff = diff - a[i];
            k = k - 1;
          
        if (diff != 0):
            print("-1");
            return;
          
    # Print the distribution
    for i in range(n):
        print(int(a[i]), end=" ");
      
# Driver code
if __name__ == '__main__':
    n, k, l, r, S, Sk = 5, 3, 1, 5, 13, 9;
  
    distribution(n, k, l, r, S, Sk);
      
# This code is contributed by PrinciRaj1992

C#




// C# implementation of the approach 
using System;
  
class GFG 
{
      
    // Function for the distribution of the number 
    static void distribution(int n, int k, int l, 
                            int r, int S, int Sk) 
    
        int []a = new int[n]; 
        int len = k, temp, rem; 
        int diff = S - Sk; 
          
        for (int i = 0; i < len; i++)
        {
              
            // Distribute the number 
            // among k elements 
            temp = Sk / k; 
            rem = Sk % k; 
            if (temp + rem >= l && temp + rem <= r)
            
                a[i] = temp; 
            
            else if (temp + rem > r)
            
                a[i] = r; 
            
            else if (temp + rem < r)
            
                Console.Write("-1"); 
                return
            
            Sk = Sk - a[i]; 
            k = k - 1; 
        
      
        // If there is some remaining 
        // sum to distribute 
        if (Sk > 0)
        
            Console.Write("-1"); 
            return
        
      
        // If there are elements remaining 
        // to distribute i.e. (n - k) 
        if (len != 0)
        
            k = n - len; 
            for (int i = len; i < n; i++) 
            {
                  
                // Divide the remaining sum into 
                // n-k elements 
                temp = diff / k; 
                rem = diff % k; 
                if (temp + rem >= l && temp + rem <= r)
                
                    a[i] = temp; 
                
                else if (temp + rem > r)
                
                    a[i] = r; 
                
                else if (temp + rem < r) 
                
                    Console.Write("-1"); 
                    return
                
                diff = diff - a[i]; 
                k = k - 1; 
            
            if (diff != 0) 
            
                Console.Write("-1"); 
                return
            
        
      
        // Print the distribution 
        for (int i = 0; i < n; i++)
        
            Console.Write(a[i] + " "); 
        
    
      
    // Driver code 
    public static void Main(String[] args)
    
        int n = 5, k = 3, l = 1, 
            r = 5, S = 13, Sk = 9; 
      
        distribution(n, k, l, r, S, Sk); 
    
}
  
// This code is contributed by PrinciRaj1992
Output:
3 3 3 2 2

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :