CountingBits.java

/*

Source: https://leetcode.com/problems/counting-bits/

5 Approaches

1st Approach (brute force)

Time: O(n log base 2 n)
Space: O(n), array of size n is required

*/

class Solution {
  public int[] countBits(int n) {

    int[] res = new int[n + 1];

    for(int i = 1; i <= n; ++i) {

      int num = i;
      int setBits = 0;

      while(num != 0) {

        if((num & 1) != 0) { // is equivalent to num % 2 != 0 to check whether the number is even or odd
          ++setBits;
        }

        num >>= 1;
      }

      res[i] = setBits;
    }

    return res;
  }
}

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/*

2nd Approach (using Brian Kernighan Algorithm)

Refer to the following link for detailed explanation of Brian Kernighan Algorithm
https://leetcode.com/problems/number-of-1-bits/discuss/1519675/JAVA-C%2B%2B-%3A-Simple-or-O(1)-Time-or-Faster-than-100-or-Detailed-Explanation

Time: O(n log base2 n)
Space: O(n), array of size n is required

*/

class Solution {
  public int[] countBits(int n) {

    int[] res = new int[n + 1];

    for(int i = 1; i <= n; ++i) {

      int num = i;
      int setBits = 0;

      while(num != 0) {
        num &= num - 1;
        ++setBits;
      }

      res[i] = setBits;
    }

    return res;
  }
}

-----------------------------------------------------------------------------------------------------------

/*

3rd Approach (more optimized than 2nd approach based on observation of the pattern formed)

A number can be either even or odd, and
Last bit of even number is 0 and odd number is 1

decimal           binary
  1     -->     1 (1 set bit)
  2     -->    10 (1 set bit)
  3     -->    11 (2 set bits)
  4     -->   100 (1 set bit)
  5     -->   101 (2 set bits)
  6     -->   110 (2 set bits)
  7     -->   111 (3 set bits)
  8     -->  1000 (1 set bit)
  9     -->  1001 (2 set bits)
 10     -->  1010 (2 set bits)

when number is even
suppose x = 6, now x / 2 = 3, here number of set bits in x(i.e. 6) are 2 and set bits in x / 2(i.e. 3) are also 2
so, we can say that when number is even, number of set bits in x = number of set bits in (x / 2)

and when number is odd
suppose x = 9, now x - 1 = 8, here number of set bits in x(i.e. 9) are 2 and set bits in x - 1(i.e. 8) are 1
so, we can say that when number is odd, number of set bits in x = number of set bits in (x - 1) + 1

Based on this observation, we can solve this problem in O(n)

Time: O(n)
Space: O(n), array of size n is required

*/

class Solution {
  public int[] countBits(int n) {

    int[] res = new int[n + 1];

    for(int i = 1; i <= n; ++i) {

      if((i & 1) == 0) { // equivalent to (i % 2 == 0) to check whether the number is even or odd
        res[i] = res[i >> 1]; // (i >> 1) is equivalent to (i / 2)
      } else {
        res[i] = res[i - 1] + 1;
      }

      res[i] = res
    }

    return res;
  }
}

-----------------------------------------------------------------------------------------------------------

/*

4th Approach (more optimized than 3rd approach also based on observation)

A number can be either even or odd, and
Last bit of even number is 0 and odd number is 1

decimal           binary
  1     -->     1 (1 set bit)
  2     -->    10 (1 set bit)
  3     -->    11 (2 set bits)
  4     -->   100 (1 set bit)
  5     -->   101 (2 set bits)
  6     -->   110 (2 set bits)
  7     -->   111 (3 set bits)
  8     -->  1000 (1 set bit)
  9     -->  1001 (2 set bits)
 10     -->  1010 (2 set bits)

when number is even
suppose x = 6, now x / 2 = 3, here number of set bits in x(i.e. 6) are 2 and set bits in x / 2(i.e. 3) are also 2
so, we can say that when number is even, number of set bits in x = number of set bits in (x / 2)

and when number is odd
suppose x = 9, now x / 2 = 4, here number of set bits in x(i.e. 9) are 2, but set bits in x / 2(i.e. 4) are 1
so, we can say that when number is odd, number of set bits in x = number of set bits in (x / 2) + 1

Based on this observation, we can solve this problem in O(n)

Time: O(n)
Space: O(n), array of size n is required

*/

class Solution {
  public int[] countBits(int n) {

    int[] res = new int[n + 1];

    for(int i = 1; i <= n; ++i) {

      if((i & 1) == 0) { // equivalent to (i % 2 == 0) to check whether the number is even or odd
        res[i] = res[i >> 1]; // (i >> 1) is equivalent to (i / 2)
      } else {
        res[i] = res[i >> 1] + 1;
      }
    }

    return res;
  }
}

-----------------------------------------------------------------------------------------------------------

/*

5th approach (more optimized version of 4th approach by avoiding if/else conditions)

Time: O(n)
Space: O(n), array of size n is required

*/

class Solution {
  public int[] countBits(int n) {

    int[] res = new int[n + 1];

    for(int i = 1; i <= n; ++i) {
      res[i] = res[i >> 1] + (i & 1);
    }

    return res;
  }
}