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Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1’s in the binary representation of i.

Clarifying Questions

1. What should be the format of the output?
   • The function should return a list of integers.
2. Are there any constraints on the value of n?
   • Yes, 0 <= n <= 10^5, as per common constraint limits for LeetCode problems.
3. Is there any specific requirement for the method to calculate the number of 1s?
   • The method should be efficient, preferably better than O(n log n), which is the time complexity of counting bits individually for each number.

Strategy

One efficient way to solve this problem is by using dynamic programming and the properties of binary numbers:

1. Binary Observations:
   • The number of 1s in a binary number can be built from the number of 1s in smaller numbers.
   • For a number i in binary: if i is an even number, the number of 1s in i is the same as the number of 1s in i // 2 (shifting right by 1 bit).
   • If i is an odd number, it has one more 1 than i - 1.
2. Dynamic Programming Approach:
   • Use an array dp where dp[i] represents the count of 1s in the binary representation of i.
   • Initialize dp[0] to 0 because the number 0 has no 1s.
   • For each i from 1 to n, calculate dp[i] based on whether i is even or odd as explained above.

Code

def countBits(n: int) -> [int]:
    # Initialize the array to store number of 1's for each integer
    dp = [0] * (n + 1)
    
    # Start from 1 and calculate number of 1's using previous numbers
    for i in range(1, n + 1):
        dp[i] = dp[i >> 1] + (i & 1)
    
    return dp

Explanation

• dp is initialized with n+1 zeros since we need counts from 0 to n.
• The loop runs from 1 to n.
  • i >> 1 is the integer division of i by 2, effectively shifting the bits to the right.
  • i & 1 checks if the least significant bit of i is 1 (i.e., whether i is odd).
  • dp[i] is calculated as the count of 1s in i >> 1 plus i & 1.
• Return the array dp which contains the result.

Time Complexity

• The time complexity is O(n) since we are iterating through each number from 0 to n once.
• The space complexity is also O(n) because we are storing the counts in a list of size n + 1.

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