Distribution of a Number in Array within a Range
Given the 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 the first element of the array will be (Sk/k) + (Sk % k), and the remaining k – 1 element 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++
#include <bits/stdc++.h>
using namespace std;
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++) {
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 (Sk > 0) {
cout << "-1" ;
return ;
}
if (len) {
k = n - len;
for ( int i = len; i < n; i++) {
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 ;
}
}
for ( int i = 0; i < n; i++) {
cout << a[i] << " " ;
}
}
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
class GFG
{
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++)
{
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 (Sk > 0 )
{
System.out.print( "-1" );
return ;
}
if (len != 0 )
{
k = n - len;
for ( int i = len; i < n; i++)
{
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 ;
}
}
for ( int i = 0 ; i < n; i++)
{
System.out.print(a[i] + " " );
}
}
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);
}
}
|
Python3
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 ):
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 (Sk > 0 ):
print ( "-1" );
return ;
if ( len ! = 0 ):
k = n - len ;
for i in range ( len , n):
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 ;
for i in range (n):
print ( int (a[i]), end = " " );
if __name__ = = '__main__' :
n, k, l, r, S, Sk = 5 , 3 , 1 , 5 , 13 , 9 ;
distribution(n, k, l, r, S, Sk);
|
C#
using System;
class GFG
{
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++)
{
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 (Sk > 0)
{
Console.Write( "-1" );
return ;
}
if (len != 0)
{
k = n - len;
for ( int i = len; i < n; i++)
{
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 ;
}
}
for ( int i = 0; i < n; i++)
{
Console.Write(a[i] + " " );
}
}
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);
}
}
|
Javascript
<script>
function distribution(n, k, l, r, S, Sk)
{
let a = new Array(n);
let len = k, temp, rem, s;
let diff = S - Sk;
for (let i = 0; i < len; i++) {
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) {
document.write( "-1" );
return ;
}
Sk = Sk - a[i];
k = k - 1;
}
if (Sk > 0) {
document.write( "-1" );
return ;
}
if (len) {
k = n - len;
for (let i = len; i < n; i++) {
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) {
document.write( "-1" );
return ;
}
diff = diff - a[i];
k = k - 1;
}
if (diff) {
document.write( "-1" );
return ;
}
}
for (let i = 0; i < n; i++) {
document.write(a[i] + " " );
}
}
let n = 5, k = 3, l = 1, r = 5, S = 13, Sk = 9;
distribution(n, k, l, r, S, Sk);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
Last Updated :
03 Feb, 2022
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