There are n servers. Each server i is currently processing a(i) amount of requests. There is another array b in which b(i) represents the number of incoming requests that are scheduled to server i. Reschedule the incoming requests in such a way that each server i holds an equal amount of requests after rescheduling. An incoming request to server i can be rescheduled only to server i-1, i, i+1. If there is no such rescheduling possible then output -1 else print number of requests hold by each server after rescheduling.
Examples:
Input : a = {6, 14, 21, 1}
b = {15, 7, 10, 10}
Output : 21
b(0) scheduled to a(0) --> a(0) = 21
b(1) scheduled to a(1) --> a(1) = 21
b(2) scheduled to a(3) --> a(3) = 11
b(3) scheduled to a(3) --> a(3) = 21
a(2) remains unchanged --> a(2) = 21
Input : a = {1, 2, 3}
b = {1, 100, 3}
Output : -1
No rescheduling will result in equal requests.Approach: Observe that each element of array b is always added to any one element of array a exactly once. Thus the sum of all elements of array b + sum of all elements of old array a = sum of all elements of new array a. Let this sum be S. Also all the elements of new array a are equal. Let each new element is x. If array a has n elements, this gives
x * n = S => x = S/n ....(1)
Thus all the equal elements of new array a is given by eqn(1). Now to make each a(i) equals to x we need to add x-a(i) to each element. We will iterate over entire array a and check whether a(i) can be made equal to x. There are multiple possibilities:
- a(i) > x: In this case a(i) can never be made equal to x. So output -1.
- a(i) + b(i) + b(i+1) = x. Simply add b(i) + b(i+1) to a(i) and update b(i), b(i+1) to zero.
- a(i) + b(i) = x. Add b(i) to a(i) and update b(i) to zero.
- a(i) + b(i+1) = x. Add b(i+1) to a(i) and update b(i+1) to zero.
After array a is completely traversed, check whether all elements of array b are zero or not. If yes then print a(0) otherwise print -1.
Why b(i) is updated to zero after addition?
Consider a test case in which b(i) is neither added to a(i-1) nor a(i). In that case, we are bounded to add b(i) to a(i+1). Thus while iterating over the array a when we begin performing computations on element a(i), first we add element b(i-1) to a(i) to take into consideration above possibility. Now if b(i-1) is already added to a(i-1) or a(i-2) then, in that case, it cannot be added to a(i). So to avoid this double addition of b(i) it is updated to zero.
The stepwise algorithm is:
1. Compute sum S and find x = S / n
2. Iterate over array a
3. for each element a(i) do:
a(i) += b(i-1)
b(i-1) = 0;
if a(i) > x:
break
else:
check for other three possibilities
and update a(i) and b(i).
4. Check whether all elements of b(i) are
zero or not.Implementation:
C++
// CPP program to schedule jobs so that// each server gets equal load.#include <bits/stdc++.h>using namespace std;
// Function to find new array aint solve(int a[], int b[], int n)
{ int i;
long long int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++) {
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0) {
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x) {
a[i] = y;
b[i] = b[i + 1] = 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x) {
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x) {
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}int main()
{ int a[] = { 6, 14, 21, 1 };
int b[] = { 15, 7, 10, 10 };
int n = sizeof(a) / sizeof(a[0]);
cout << solve(a, b, n);
return 0;
} |
Java
// Java program to schedule jobs so that// each server gets equal load.class GFG
{// Function to find new array astatic int solve(int a[], int b[], int n)
{ int i;
int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++)
{
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0)
{
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x)
{
a[i] = y;
b[i]= 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x)
{
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x)
{
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}// Driver codepublic static void main(String[] args)
{ int a[] = { 6, 14, 21, 1 };
int b[] = { 15, 7, 10, 10 };
int n = a.length;
System.out.println(solve(a, b, n));
}} // This code contributed by Rajput-Ji |
Python3
# Python3 program to schedule jobs so that # each server gets an equal load. # Function to find new array a def solve(a, b, n):
s = 0
# find sum S of both arrays a and b.
for i in range(0, n):
s += a[i] + b[i]
# Single element case.
if n == 1:
return a[0] + b[0]
# This checks whether sum s can be divided
# equally between all array elements. i.e.
# whether all elements can take equal value
# or not.
if s % n != 0:
return -1
# Compute possible value of new
# array elements.
x = s // n
for i in range(0, n):
# Possibility 1
if a[i] > x:
return -1 # ensuring that all elements of
# array b are used.
if i > 0:
a[i] += b[i - 1]
b[i - 1] = 0
# If a(i) already updated to x
# move to next element in array a.
if a[i] == x:
continue
# Possibility 2
y = a[i] + b[i]
if i + 1 < n:
y += b[i + 1]
if y == x:
a[i] = y
b[i] = 0
if i + 1 < n: b[i + 1] = 0
continue
# Possibility 3
if a[i] + b[i] == x:
a[i] += b[i]
b[i] = 0
continue
# Possibility 4
if i + 1 < n and a[i] + b[i + 1] == x:
a[i] += b[i + 1]
b[i + 1] = 0
continue
# If a(i) can not be made equal
# to x even after adding all
# possible elements from b(i)
# then print -1.
return -1
# check whether all elements of b
# are used.
for i in range(0, n):
if b[i] != 0:
return -1 # Return the new array element value.
return x
# Driver Codeif __name__ == "__main__":
a = [6, 14, 21, 1]
b = [15, 7, 10, 10]
n = len(a)
print(solve(a, b, n))
# This code is contributed by Rituraj Jain |
C#
// C# program to schedule jobs so that // each server gets equal load. using System;
class GFG
{ // Function to find new array a static int solve(int []a, int []b, int n)
{ int i;
int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++)
{
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0)
{
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x)
{
a[i] = y;
b[i]= 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x)
{
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x)
{
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
} // Driver code public static void Main(String[] args)
{ int []a = { 6, 14, 21, 1 };
int []b = { 15, 7, 10, 10 };
int n = a.Length;
Console.WriteLine(solve(a, b, n));
} } // This code has been contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to schedule jobs so that// each server gets equal load. // Function to find new array a
function solve(a, b, n)
{ let i;
let s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
let x = s / n;
for (i = 0; i < n; i++)
{
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0)
{
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
let y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x)
{
a[i] = y;
b[i]= 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x)
{
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x)
{
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}// Driver Code let a = [6, 14, 21, 1];
let b = [15, 7, 10, 10];
let n = a.length;
document.write(solve(a, b, n));
// This code is contributed by avijitmondal1998.</script> |
Output
21
Time Complexity: O(n)
Auxiliary Space : O(1) If we are not allowed to modify original arrays, then O(n)
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