Help me in solving EQUALELE problem

help
,
1

My issue

Can anyone pls help me in resolving this issue for Time Limit exceeded in Equal Elements problem

My code

#include <stdio.h>

int main(void) {
	// your code goes here
	int t,N,count,grt,eql,temp;
	scanf("%d",&t);
	while(t--)
	{
	count = 0;
	temp = 0;
	eql = 0;
	scanf("%d",&N);
	int a[N],b[N];
	for(int i=0;i <N; i++)
	{
	scanf("%d",&a[i]);
	}
	for(int i=0; i<N; i++)
	{
	for(int j=0; j<N; j++)
	{
	if(a[i]==a[j])
	{
	count = count + 1;
	}
	}
    b[i] = count;
	count = 0;
    }	
    
    // all not equal
    for(int i=0; i<N; i++)
    {
    if(b[i] == 1)
    {
    temp = temp + 1;
    }
	}

	// all equal
	for(int j=0; j<N; j++)
    {
    if(b[j] == N)
    {
    eql = eql + 1;
    }
	}

	// some equal
	for(int k=0; k<N; k++)
	{
    if(b[0] < b[k])
    {
    b[0] = b[k];    
    }
	}
	grt = b[0];
	
	if(temp == N)
	{
	printf("%d \n",N-1);
	}
	
	else if(eql == N)
	{
	printf("0 \n");
	}
	
	else
	{
	printf("%d \n", N-grt);    
	}
	}
	return 0;
}

Learning course: Arrays, Strings & Sorting
Problem Link: CodeChef: Practical coding for everyone

2

@sriharijosyula
U can reduce the time complexity by sorting it and counting just the adjacent numbers for maximum frequency element.
This will reduce your time complexity from O(N^2) to O(NlogN) and thus will remove your tle.

3

So, by sorting through merge sort algorithm will the time complexity reduce here in this case?

Thanks for your kind reply

4

@sriharijosyula
Yeah it will work fine