# One more approach Using a boolean array (22ms)

• In this solution, a different way to iterate over the elements is given. We keep track of how many sides of a square are formed and what elements are traversed. When a new side is formed, we will again start iterating from zeroth element. I think, this solution avoids the same combinations being selected again, which is the case in other solutions using an sum array of size 4. Repeating again from zeroth element looks like a bad approach but we are doing that only when a side is formed.

For example, given [2,1,1,2,3,3,4], we know subset [2,1,1] is a bad choice but this will be chosen 4 times in solutions which use sum array. This solution avoids that.

``````public class Solution {
public boolean makesquare(int[] nums) {
int sum = 0, max = 0;
for(int i = 0; i < nums.length; i++) {
if(nums[i] < 0) return false;
sum += nums[i];
max = Math.max(max, nums[i]);
}
if(sum % 4 != 0 || max > sum/4) return false;
Arrays.sort(nums);
return helper(nums, new boolean[nums.length], sum/4, 0, 0, 0);
}

boolean helper(int nums[], boolean bool[], int side, int no_formed, int index, int curr_sum) {

if(curr_sum == side && no_formed == 2) return true;
if(index >= nums.length || curr_sum > side) return false;

// when a new side formed
if(curr_sum == side) {
no_formed++;
index = 0;
curr_sum = 0;
}

for(int i = index; i < nums.length; i++) {
if(!bool[i]) {
if(curr_sum + nums[i] > side) return false;
bool[i] = true;
if(helper(nums, bool, side, no_formed, i+1, curr_sum + nums[i])) return true;
bool[i] = false;
}
}
return false;
}
}
``````

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