Count K-Reducible Numbers Less Than N - Practice Coding Problems

Count K-Reducible Numbers Less Than N - Problem

You are given a binary string s representing a number n in its binary form. You are also given an integer k.

An integer x is called k-reducible if performing the following operation at most k times reduces it to 1:

Replace x with the count of set bits in its binary representation.

For example, the binary representation of 6 is "110". Applying the operation once reduces it to 2 (since "110" has two set bits). Applying the operation again to 2 (binary "10") reduces it to 1 (since "10" has one set bit).

Return an integer denoting the number of positive integers less than n that are k-reducible. Since the answer may be too large, return it modulo 10⁹ + 7.

Input & Output

Example 1 — Basic Case

$ Input: s = "111", k = 1

› Output: 3

💡 Note: n = 7 in decimal. Check numbers 1-6: 1 (already 1, 0 steps), 2 (binary "10", 1 bit, 1 step), 3 (binary "11", 2 bits → 1, 1 step), 4 (binary "100", 1 bit, 1 step), 5 (binary "101", 2 bits → 1, 1 step), 6 (binary "110", 2 bits → 1, 1 step). Numbers 1,2,3,4,5,6 all need ≤1 step, but we need exactly the count. Actually: 1(0 steps), 2(1 step: 2→1), 3(2 steps: 3→2→1). So 1,2 are 1-reducible. Wait, let me recalculate: 3 has 2 bits, 2→1 is 1 step. So 1,2,3 are all 1-reducible. Answer is 3.

Example 2 — Larger k

$ Input: s = "1000", k = 2

› Output: 6

💡 Note: n = 8 in decimal. Check 1-7: All numbers 1-7 can be reduced to 1 in at most 2 steps, except maybe some. Most small numbers are 2-reducible, so answer is 6.

Example 3 — Small Case

$ Input: s = "11", k = 1

› Output: 2

💡 Note: n = 3 in decimal. Check 1,2: 1 needs 0 steps, 2 needs 1 step (2→1). Both are 1-reducible.

Constraints

1 ≤ s.length ≤ 10⁵
s consists of only '0' and '1'
s does not have leading zeros
1 ≤ k ≤ 5

Visualization

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Understanding the Visualization

Input

Binary string s="111" (n=7), k=1

Process

Check each number 1-6 for k-reducibility

Output

Count of k-reducible numbers: 3

Key Takeaway

🎯 Key Insight: Precompute reduction steps for bit counts, then use digit DP to count numbers with k-reducible bit counts efficiently

Asked in

G Google 15 f Meta 12

The key insight is that most numbers reduce to 1 very quickly, so we can precompute reduction steps for all possible bit counts and use digit DP to count efficiently. The memoized approach with digit DP is optimal: Time: O(n × b), Space: O(n × b)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n × log n)	O(1)	Check each number from 1 to n-1 by simulating the reduction process
Memoized Reduction with Digit DP	O(n × b)	O(n × b)	Precompute reduction steps and use digit DP to count efficiently

Brute Force Simulation — Algorithm Steps

Convert binary string s to decimal n
For each number from 1 to n-1, simulate reduction
Count set bits repeatedly until reaching 1
Count numbers that reduce in ≤ k steps

Visualization

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Step-by-Step Walkthrough

Convert Input

Convert binary string to decimal

Test Each Number

For each i from 1 to n-1, simulate reduction

Count Valid

Count numbers that reduce in ≤ k steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int countBits(long long x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

int reductionSteps(long long x, int maxK) {
    int steps = 0;
    while (x != 1) {
        x = countBits(x);
        steps++;
        if (steps > maxK) return steps;
    }
    return steps;
}

int solution(char* s, int k) {
    const int MOD = 1000000007;
    long long n = 0;
    for (int i = 0; s[i]; i++) {
        n = n * 2 + (s[i] - '0');
    }
    
    int count = 0;
    for (long long i = 1; i < n; i++) {
        if (reductionSteps(i, k) <= k) {
            count++;
        }
    }
    
    return count % MOD;
}

int main() {
    char s[1000];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × log n)

Check n numbers, each takes O(log n) steps to reduce

⚡ Linearithmic

Space Complexity

O(1)

Only using variables for counting

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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