class Solution {
public int[] countOppositeParity(int[] nums) {
// approach 1 - checks every pair
int n = nums.length;
// i dont get the logic of how the conditions satisfy each other, and score part.
int[] ans = new int[n];
for(int i=0; i<n;i++) {
int score = 0;
for(int j=i+1; j<n; j++) {
// check if value parity is opp
boolean oppositeValue = (nums[i]%2!=nums[j]%2);
if(oppositeValue) {
score++;
}
}
ans[i] = score;
}
return ans;
}
}
// approach 2 - checks every element once
oddCount =0;
evenCount=0;
for(int =n-1; i>=0; i--) {
if(n%2==0) {
// curr->even, it pairs with all odds to its right
ans[i] = evenCount;
oddCount++;
} else {
//curr-> odd, it pairs with all even to its right
ans[i] = oddCount;
evenCount++;
}
}
