Count of operations to make all elements of array a[] equal to its min element by performing a[i] – b[i]
Last Updated :
30 Dec, 2022
Given two array a[] and b[] of size N, the task is to print the count of operations required to make all the elements of array a[i] equal to its minimum element by performing a[i] – b[i] where its always a[i] >= b[i]. If it is not possible then return -1.
Example:
Input: a[] = {5, 7, 10, 5, 15} b[] = {2, 2, 1, 3, 5}
Output: 8
Explanation:
Input array is a[] = 5, 7, 10, 5, 15 and b[] = 2, 2, 1, 3, 5. The minimum from a[] is 5.
Now for a[0] we don’t have to perform any operation since its already 5.
For i = 1, a[1] – b[1] = 7 – 2 = 5. (1 operation)
For i = 2, a[2] – b[2] = 10 – 1 = 9 – 1 = 8 – 1 = 7 – 1 = 6 – 1 = 5 (5 operation)
For i = 3, a[3] = 5
For i = 4, a[4] – b[4] = 15 – 5= 10 – 5 = 5 (2 operation)
The total number of operations required is 8.
Input: a[] = {1, 3, 2} b[] = {2, 3, 2}
Output:-1
Explanation:
It is not possible to convert the array a[] into equal elements.
Approach: To solve the problem mentioned above follow the steps given below:
- Find minimum from array a[]. Initialize a variable ans = -1 that stores resultant subtractions operation.
- Iterate from minimum element of array a[] to 0 and initialize variable curr to 0 that stores the count subtraction to make the array element equal.
- Traverse in the array and check if a[i] is not equal to x which is the minimum element in the first array, then make it equal to minimum else update curr equal to zero.
- Check if curr is not equal to 0 then update ans as curr finally return the ans.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findMinSub( int a[], int b[], int n)
{
int min = INT_MAX;
for ( int i = 0; i < n; i++) {
if (a[i] < min)
min = a[i];
}
int ans = -1;
for ( int x = min; x >= 0; x--)
{
int curr = 0;
for ( int i = 0; i < n; i++) {
if (a[i] != x) {
if (b[i] > 0
&& (a[i] - x) % b[i] == 0) {
curr += (a[i] - x) / b[i];
}
else {
curr = 0;
break ;
}
}
}
if (curr != 0) {
ans = curr;
break ;
}
}
return ans;
}
int main()
{
int a[] = { 5, 7, 10, 5, 15 };
int b[] = { 2, 2, 1, 3, 5 };
int n = sizeof (a) / sizeof (a[0]);
cout << findMinSub(a, b, n);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int findMinSub( int a[], int b[], int n)
{
int min = Integer.MAX_VALUE;
for ( int i = 0 ; i < n; i++)
{
if (a[i] < min)
min = a[i];
}
int ans = - 1 ;
for ( int x = min; x >= 0 ; x--)
{
int curr = 0 ;
for ( int i = 0 ; i < n; i++)
{
if (a[i] != x)
{
if (b[i] > 0 &&
(a[i] - x) % b[i] == 0 )
{
curr += (a[i] - x) / b[i];
}
else
{
curr = 0 ;
break ;
}
}
}
if (curr != 0 )
{
ans = curr;
break ;
}
}
return ans;
}
public static void main(String[] args)
{
int a[] = { 5 , 7 , 10 , 5 , 15 };
int b[] = { 2 , 2 , 1 , 3 , 5 };
int n = a.length;
System.out.print(findMinSub(a, b, n));
}
}
|
Python3
def findMinSub(a, b, n):
min = a[ 0 ]
for i in range ( 0 , n):
if a[i] < min :
min = a[i]
ans = - 1
for x in range ( min , - 1 , - 1 ):
curr = 0
for i in range ( 0 , n):
if a[i] ! = x:
if (b[i] > 0 and
(a[i] - x) % b[i] = = 0 ):
curr + = (a[i] - x) / / b[i]
else :
curr = 0
break
if curr ! = 0 :
ans = curr
break
return ans
a = [ 5 , 7 , 10 , 5 , 15 ]
b = [ 2 , 2 , 1 , 3 , 5 ]
n = len (a)
print (findMinSub(a, b, n))
|
C#
using System;
class GFG{
static int findMinSub( int []a, int []b, int n)
{
int min = Int32.MaxValue;
for ( int i = 0; i < n; i++)
{
if (a[i] < min)
min = a[i];
}
int ans = -1;
for ( int x = min; x >= 0; x--)
{
int curr = 0;
for ( int i = 0; i < n; i++)
{
if (a[i] != x)
{
if (b[i] > 0 &&
(a[i] - x) % b[i] == 0)
{
curr += (a[i] - x) / b[i];
}
else
{
curr = 0;
break ;
}
}
}
if (curr != 0)
{
ans = curr;
break ;
}
}
return ans;
}
public static void Main()
{
int []a = { 5, 7, 10, 5, 15 };
int []b = { 2, 2, 1, 3, 5 };
int n = a.Length;
Console.Write(findMinSub(a, b, n));
}
}
|
Javascript
<script>
function findMinSub(a , b , n) {
var min = Number.MAX_VALUE;
for (i = 0; i < n; i++) {
if (a[i] < min)
min = a[i];
}
var ans = -1;
for (x = min; x >= 0; x--) {
var curr = 0;
for (i = 0; i < n; i++) {
if (a[i] != x) {
if (b[i] > 0 && (a[i] - x) % b[i] == 0) {
curr += (a[i] - x) / b[i];
} else {
curr = 0;
break ;
}
}
}
if (curr != 0) {
ans = curr;
break ;
}
}
return ans;
}
var a = [ 5, 7, 10, 5, 15 ];
var b = [ 2, 2, 1, 3, 5 ];
var n = a.length;
document.write(findMinSub(a, b, n));
</script>
|
Time Complexity: O(min*n) // min is the minimum element in the array
Auxiliary Space: O(1)
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