Optimizing C code [closed] - c++

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Suppose we have an array of numbers say {1,2,3} and we want to equalize the numbers in the least number of turns possible; where the definition of a "turn" is as follows:
In a turn, you need to fix the value of one of the elements as is, and increment every other number by 1.
Considering the eg. already mentioned - A={1,2,3} , the goal is to equalize them.What I've already done is formulate the logic i.e The method to using a minimum number of turns is to choose the maximum number in each turn.
Iteration 1: Hold A[2]=3. Array at end of iteration => {2,3,3}
Iteration 2: Hold A[2]=3. Array at end of iteration => {3,4,3}
Iteration 3: Hold A[1]=4. Array at end of iteration => {4,4,4}
So,number of turns taken = 3
The code I've written is as follows:
#include<iostream>
#include<stdio.h>
int findMax(int *a,int n)
{
int i,max;
max=1;
for(i=2;i<=n;i++)
{
if(a[i]>a[max])
{
max=i;
}
}
return max;
}
int equality(int *a,int n)
{
int i;
for(i=1;i<n;i++)
{
if(a[i]!=a[i+1]) return 0;
}
return 1;
}
int main()
{
int a[100],i,count,t,posn_max,n,ip=0;
scanf("%d",&t);
while(ip<t)
{
count=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
while(equality(a,n)==0)
{
posn_max=findMax(a,n);
for(i=1;i<=n;i++)
{
if(i!=posn_max)
{
a[i]=a[i]+1;
}
}
count++;
}
printf("%d\n",count);
ip++;
}
return 0;
}
This gives me the correct answer I need alright. But I want to optimize it further.
My Time Limit is 1.0 s . But the judge site tells me my code takes 1.01s. Can anyone help me out?
As far as I can see, I've used scanf/printf statements as compared to cout/cin, in a bid to optimize the input/output part. But what else should I be doing better?

Answer 1

In your algorithm, you are increasing all numbers in the expect for the maximum.
If you do it the other way around, decreasing the maximum and leaving the rest of the numbers, the result should be the same (but with much less memory/array operations)!
To make it even faster, you can get rid of the memory oeprations completely (as suggested by Ivaylo Strandjev also): Find the minimum number and by the idea above (of decreasing numbers instead of increasing) you know how much decreases you require to decrease all numbers to this minimum number. So, after finding the minimum you need one loop to calculate the number of turns.
Take your example of {1,2,3}
The minimum is 1
Number of turns: (1-1)+(2-1)+(3-1) = 0 + 1 + 2 = 3
If you are really clever, it is possible to calculate the number of turns directly when inputting the numbers and keeping track of the current minimum number... Try it! ;)

Answer 2

You only care about the count not about the actual actions you need to perform. So instead of performing the moves one by one try to find a way to count the number of moves without performing them. The code you wrote will not pass in the time limit no matter how well you optimize it. The maximum element observation you've made will help you along the way.

Answer 3

Besides the other comments, if I get this right and your code is just a little bit too slow, here are two optimizations which should help you.
First, you can combine equality() and findMax() and only scan once through the array instead of your current worst case (twice).
Second, you can split the "increase" loop into two parts (below and above the max position). This will remove the effort to check the position in the loop.

Answer 4

1) Try unrolling the loops
2) Can you use SIMD instruction? That would really speed this code up

Answer 5

I would printf in a separate thread, since it's an I/O operation and is much slower than your calculations.
It also does not demand complicated management e.g Producer-Consumer queue, since you only pass the ordered numbers from 0 to last count.
Here's the pseudo-code:
volatile int m_count = 0;
volatile bool isExit = false;
void ParallelPrint()
{
int currCount = 0;
while (!isExit)
{
while (currCount < m_count)
{
currCount++;
printf("%d\n", currCount);
}
Sleep(0); // just content switch
}
}
Open the thread before the scanf("%d",&t); (I guess this initialization time is not counted), and close the thread by isExit = true; before the return from your Main().