Given two positive integer L and R, the task is to count the total number of set bits in the binary representation of all the numbers from L to R.
Examples:
Input:_ L = 3, R = 5 _
Output:_ 5 _
Explanation:_ (3)10 = (11)2, (4)10 = (100)2, (5)10 = (101)2 _
So, Total set bit in range 3 to 5 is 5
Input:_ L = 10, R = 15 _
Output:_ 17 _
Method 1 – Naive Approach: The idea is to run a loop from L to R and sum the count of set bits in all numbers from L to R.
Below is the implementation of the above approach:
- C++
// C++ program for the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function to count set bits in x
unsigned **int**
countSetBitsUtil(unsigned **int** x)
{
// Base Case
**if** (x <= 0)
**return** 0;
// Recursive Call
**return** ((x % 2 == 0 ? 0 : 1)
+ countSetBitsUtil(x / 2));
}
// Function that returns count of set bits
// present in all numbers from 1 to N
unsigned **int** countSetBits(unsigned **int** L,
unsigned **int** R)
{
// Initialize the result
**int** bitCount = 0;
**for** (``**int** i = L; i <= R; i++) {
bitCount += countSetBitsUtil(i);
}
// Return the setbit count
**return** bitCount;
}
// Driver Code
**int** main()
{
// Given L and R
**int** L = 3, R = 5;
// Function Call
**printf**``(``"Total set bit count is %d"``,
countSetBits(L, R));
**return** 0;
}
Output:
Total set bit count is 5
Time Complexity:_ O(N*Log N) _
Auxiliary Space:_ O(1)_
Method 2 – Better Approach: The idea is to observe bits from rightmost side at distance i than bits get inverted after 2i position in vertical sequence.
For Example:
L = 3, R = 5
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
Observe the right most bit (i = 0) the bits get flipped after (20 = 1)
Observe the 3rd rightmost bit (i = 2) the bits get flipped after (22 = 4).
Therefore, We can count bits in vertical fashion such that at i’th right most position bits will be get flipped after 2i iteration.
Below is the implementation of the above approach:
- C++
// C++ program for the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function that counts the set bits
// from 0 to N
**int** countSetBit(``**int** n)
{
**int** i = 0;
// To store sum of set bits from 0 - N
**int** ans = 0;
// Untill n >= to 2^i
**while** ((1 << i) <= n) {
// This k will get flipped after
// 2^i iterations
**bool** k = 0;
// Change is iterator from 2^i to 1
**int** change = 1 << i;
// This will loop from 0 to n for
// every bit position
**for** (``**int** j = 0; j <= n; j++) {
ans += k;
**if** (change == 1) {
// When change = 1 flip the bit
k = !k;
// Again set change to 2^i
change = 1 << i;
}
**else** {
change--;
}
}
// Increment the position
i++;
}
**return** ans;
}
// Function that counts the set bit
// in the range (L, R)
**int** countSetBits(``**int** L, **int** R)
{
// Return the count
**return** **abs**``(countSetBit(R)
- countSetBit(L - 1));
}
// Driver Code
**int** main()
{
// Given L and R
**int** L = 3, R = 5;
// Function Call
cout << "Total set bit count is "
<< countSetBits(L, R) << endl;
**return** 0;
}
Output:
Total set bit count is 5
Time Complexity:_ O((L + R)*K), where K is the number of bits in L and R. _
Auxiliary Space:_ O(1)_
Method 3 – Tricky If the input number is of the form 2b – 1 e.g., 1, 3, 7, 15, … etc, the number of set bits is b * 2(b-1). This is because for all the numbers 0 to 2b – 1, if you complement and flip the list you end up with the same list (half the bits are set and half bits are unset).
If the number does not have all set bits, then let m is the position of leftmost set bit. The number of set bits in that position is n – (1 << m) + 1. The remaining set bits are in two parts:
- The bits in the (m – 1) positions down to the point where the leftmost bit becomes 0
- The 2(m – 1) numbers below that point, which is the closed form above.
For Example: N = 6
0|0 0
0|0 1
0|1 0
0|1 1
-|--
1|0 0
1|0 1
1|1 0
From the above we have:
- The leftmost set bit is in position 2 (positions are considered starting from 0).
- If we mask that off what remains is 2 (the “1 0” in the right part of the last row), So the number of bits in the 2nd position (the lower left box) is 3 (that is, 2 + 1).
- The set bits from 0-3 (the upper right box above) is 2*2(2 – 1) = 4.
- The box in the lower right is the remaining bits we haven’t yet counted, and is the number of set bits for all the numbers up to 2 (the value of the last entry in the lower right box) which can be figured recursively.
Below is the implementation of the above approach:
- C++
// C++ program for the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
unsigned **int** countSetBit(unsigned **int** n);
// Returns position of leftmost set bit
// The rightmost position is taken as 0
unsigned **int** getLeftmostBit(``**int** n)
{
**int** m = 0;
**while** (n > 1) {
n = n >> 1;
m++;
}
**return** m;
}
// Function that gives the position of
// previous leftmost set bit in n
unsigned **int** getNextLeftmostBit(``**int** n, **int** m)
{
unsigned **int** temp = 1 << m;
**while** (n < temp) {
temp = temp >> 1;
m--;
}
**return** m;
}
// Function to count the set bits between
// the two numbers N and M
unsigned **int** _countSetBit(unsigned **int** n,
**int** m)
{
// Base Case
**if** (n == 0)
**return** 0;
// Get position of next leftmost set bit
m = getNextLeftmostBit(n, m);
// If n is of the form 2^x-1
**if** (n == ((unsigned **int**``)1 << (m + 1)) - 1)
**return** (unsigned **int**``)(m + 1) * (1 << m);
// Update n for next recursive call
n = n - (1 << m);
**return** ((n + 1)
+ countSetBit(n)
+ m * (1 << (m - 1)));
}
// Function that returns count of set
// bits present in all numbers from 1 to n
unsigned **int** countSetBit(unsigned **int** n)
{
// Get the position of leftmost set
// bit in n
**int** m = getLeftmostBit(n);
// Use the position
**return** _countSetBit(n, m);
}
// Function that counts the set bits
// between L and R
**int** countSetBits(``**int** L, **int** R)
{
**return** **abs**``(countSetBit(R)
- countSetBit(L - 1));
}
// Driver Code
**int** main()
{
// Given L and R
**int** L = 3, R = 5;
// Function Call
cout << "Total set bit count is "
<< countSetBits(L, R);
**return** 0;
}
Output:
Total set bit count is 5
Time Complexity:_ O(log N) _
Auxiliary Space:_ O(1)_
Method 4 – using setbit: In setbit method count one by one set bit of each number in range L to R using last bit, check to last bit and if it is set then increase the count and finally sum up over it.
Below is the implementation of the above approach:
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