Sum of Values at Indices With K Set Bits | Leetcode 2859

 2859. sum of Values at Indices With K Set Bits | Leetcode weekly 363 

You are given a 0-indexed integer array nums and an integer k.

Return an integer that denotes the sum of elements in nums whose corresponding indices have exactly k set bits in their binary representation.

The set bits in an integer are the 1's present when it is written in binary.

• For example, the binary representation of 21 is 10101, which has 3 set bits.

 

Example 1:

Input: nums = [5,10,1,5,2], k = 1
Output: 13
Explanation: The binary representation of the indices are: 
0 = 0002
1 = 0012
2 = 0102
3 = 0112
4 = 1002 
Indices 1, 2, and 4 have k = 1 set bits in their binary representation.
Hence, the answer is nums[1] + nums[2] + nums[4] = 13.

Example 2:

Input: nums = [4,3,2,1], k = 2
Output: 1
Explanation: The binary representation of the indices are:
0 = 002
1 = 012
2 = 102
3 = 112
Only index 3 has k = 2 set bits in its binary representation.
Hence, the answer is nums[3] = 1.

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 105
• 0 <= k <= 10

Solution : 

The logic behind the function findsetbits() is that 

• Even number has the same number of set bits as their half . for example 4 has the same number of sets bits as 4/2 = 2  or also  findsetbits(6) = findsetbits(3).
• 
  
• Odd number has one plus the number of set bits in their half. so findsetbits(5) = findsetbits(2) + 1.

class Solution { public:     int findsetbits(int a ,vector<int>&dp )     {         if( a == 0)         {             return 0;         }         if(a == 1)         {             return 1;         }         if(dp[a] != -1)         {             return dp[a];         }         return  dp[a] = (a%2== 0) ? (findsetbits(a/2 , dp)) : findsetbits(a/2 , dp)+1;     }     int sumIndicesWithKSetBits(vector<int>& nums, int k) {               vector<int>dp(nums.size()+1 , -1 );         int sum = 0;         for(int i =0 ; i < nums.size() ; i++)         {              int x = findsetbits(i , dp);             cout << x << "\n";             if(x == k)             {                 sum += nums[i];             }         }         return sum;           } };