Accepted java solution

  • 1
    • Iterate over all the numbers with n bits, that is, let's iterate over 0 <= i < 2^n and look at each i number as a combination. A combination (number) is valid if bit count equals k.
    • If count of 1bits equal k then add new list/combination to the result (each number in the new list/combination is derived from the position of the corresponding 1 bit)
        public List<List<Integer>> combine(int n, int k) {
            List<List<Integer>> res = new ArrayList<>();
            long num = 1 << n;
            for (int i=0; i<num; i++) {
                int curr = i;
                int bit1Count = 0;
                while (curr > 0) {
                    curr = curr & curr-1;
                    if (bit1Count > k) {
                if (bit1Count == k) {
                    add(n, i, res);    
            return res;   
        private void add(int n, int curr, List<List<Integer>> res) {
            List<Integer> l = new LinkedList<>();
            for (int b=0; b<n; b++) {
                if (((curr >> b) & 1) == 1){

  • 0

    Awesome! I think your idea can combine DP to get faster result. Actually you tried every possible combinations. If using DP. you don't need to worry about these impossible combinations.

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