Number of Good pairs in O(n)

// can someone please explain what's happening inside the for loop

public int numIdenticalPairs(int[] A) {
    int res = 0, count[] = new int[101];
    for (int a: A) {
        res += count[a]++;
    }
    return res;
}

You iterate over an array of interger which work as your indices in the new generated count[]. The result sums up all the values in the "count" array that where visited. So if your input array has only unique numbers the result will always be 0. Also your inputArray must not contain more than 101 unique values otherwise your index is out of bounds.

e.g.
input array: [0,1,2] -> result=0
input array: [1,1,2] -> result=1
...

the more often a value is in your input-array the bigger inpact this value has on your result.

You use post increment, this mean you return the value of count[a] to res and then increment the value.
So res will have the sim of count values before the increment but the values in count will increase by 1

The counting of pairs between identical values forms a mathematical series. Lets consider each identical value as a node in a node grid containing each other identical value. The math works out that for each additional node added to the grid, it will add the same number of pairs to the grid as the number of nodes already in the grid (since it can pair with every node already in the grid).

0 Node in Grid
+1 Node 
Number of Pairs: 0

1 Node in Grid
+1 Node
Number of Pairs: 1 = (0 + 1)

2 Node in Grid
+1 Node
Number of Pairs: 3 = (0 + 1 + 2)

3 Node in Grid
+1 Node
Number of Pairs: 6 = (0 + 1 + 2 + 3)

And so on. The formula you have above is keeping track of this increasing node count of each identical value. It's then also adding this node count to the total sum res variable. So your loop is adding the current count of identical values, then incrementing that count of identical values for the next time it gets another identical value.