Read the problem statement clarification at the top of the discussion post. You are misinterpreting the question, observe the table given carefully.

Is that supposed to be a good thing? :3

Yes I did think of a way without actually sorting. I coded it and submitted the solution, but for some reason I'm getting a few test cases wrong. Could you have a look at it?

Thanks!

#include <bits/stdc++.h>
using namespace std;
 
bool pyramid(int m, int* count)
{
	int blocks = 0;
	for(int i = 1000000; i > m; i--)
		blocks += count[i];		//counting number of blocks with side > m
	for(int i = m; i > 0; i--)
	{
		if(count[i] == 0) 
			blocks--;		//if for any side i(<m) the block is not present, we use one of the previous unused blocks
		else
			blocks += (count[i] - 1);		//else we add the extra blocks
 
		if(blocks < 0)
			return false;		//if we ever run out of blocks, return false

The problem can be solved simply using sorting, but I think using binary search to solve it would be much more fun and informative. What do you all think?