LeetCode #2842 — HARD

Count K-Subsequences of a String With Maximum Beauty

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s and an integer k.

A k-subsequence is a subsequence of s, having length k, and all its characters are unique, i.e., every character occurs once.

Let f(c) denote the number of times the character c occurs in s.

The beauty of a k-subsequence is the sum of f(c) for every character c in the k-subsequence.

For example, consider s = "abbbdd" and k = 2:

f('a') = 1, f('b') = 3, f('d') = 2
Some k-subsequences of s are:
- "abbbdd" -> "ab" having a beauty of f('a') + f('b') = 4
- "abbbdd" -> "ad" having a beauty of f('a') + f('d') = 3
- "abbbdd" -> "bd" having a beauty of f('b') + f('d') = 5

Return an integer denoting the number of k-subsequences whose beauty is the maximum among all k-subsequences. Since the answer may be too large, return it modulo 10⁹ + 7.

A subsequence of a string is a new string formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

Notes

f(c) is the number of times a character c occurs in s, not a k-subsequence.
Two k-subsequences are considered different if one is formed by an index that is not present in the other. So, two k-subsequences may form the same string.

Example 1:

Input: s = "bcca", k = 2
Output: 4
Explanation: From s we have f('a') = 1, f('b') = 1, and f('c') = 2.
The k-subsequences of s are: 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('a') = 2 
bcca having a beauty of f('c') + f('a') = 3
bcca having a beauty of f('c') + f('a') = 3 
There are 4 k-subsequences that have the maximum beauty, 3. 
Hence, the answer is 4.

Example 2:

Input: s = "abbcd", k = 4
Output: 2
Explanation: From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. 
The k-subsequences of s are: 
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5 
There are 2 k-subsequences that have the maximum beauty, 5. 
Hence, the answer is 2.

Constraints:

1 <= s.length <= 2 * 10⁵
1 <= k <= s.length
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s and an integer k. A k-subsequence is a subsequence of s, having length k, and all its characters are unique, i.e., every character occurs once. Let f(c) denote the number of times the character c occurs in s. The beauty of a k-subsequence is the sum of f(c) for every character c in the k-subsequence. For example, consider s = "abbbdd" and k = 2: f('a') = 1, f('b') = 3, f('d') = 2 Some k-subsequences of s are: "abbbdd" -> "ab" having a beauty of f('a') + f('b') = 4 "abbbdd" -> "ad" having a beauty of f('a') + f('d') = 3 "abbbdd" -> "bd" having a beauty of f('b') + f('d') = 5 Return an integer denoting the number of k-subsequences whose beauty is the maximum among all k-subsequences. Since the answer may be too large, return it modulo 109 + 7. A subsequence of a string is a new string formed from the original string by deleting some (possibly none) of the

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math · Greedy

Example 1

"bcca"
2

Example 2

"abbcd"
4

Core Insight

What unlocks the optimal approach

Since every character appears once in a k-subsequence, we can solve the following problem first: Find the total number of ways to select <code>k</code> characters such that the sum of their frequencies is maximum.
An obvious case to eliminate is if <code>k</code> is greater than the number of distinct characters in <code>s</code>, then the answer is <code>0</code>.
We are now interested in the top frequencies among the characters. Using a map data structure, let <code>cnt[x]</code> denote the number of characters that have a frequency of <code>x</code>.
Starting from the maximum value <code>x</code> in <code>cnt</code>. Let <code>i = min(k, cnt[x])</code> we add to our result <code> cnt[x]Ci * xi</code> representing the number of ways to select <code>i</code> characters from all characters with frequency <code>x</code>, multiplied by the number of ways to choose each individual character. Subtract <code>i</code> from <code>k</code> and continue downwards to the next maximum value.
Powers, combinations, and additions should be done modulo <code>109 + 7</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
class Solution {
    private final int mod = (int) 1e9 + 7;

    public int countKSubsequencesWithMaxBeauty(String s, int k) {
        int[] f = new int[26];
        int n = s.length();
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (++f[s.charAt(i) - 'a'] == 1) {
                ++cnt;
            }
        }
        if (cnt < k) {
            return 0;
        }
        Integer[] vs = new Integer[cnt];
        for (int i = 0, j = 0; i < 26; ++i) {
            if (f[i] > 0) {
                vs[j++] = f[i];
            }
        }
        Arrays.sort(vs, (a, b) -> b - a);
        long ans = 1;
        int val = vs[k - 1];
        int x = 0;
        for (int v : vs) {
            if (v == val) {
                ++x;
            }
        }
        for (int v : vs) {
            if (v == val) {
                break;
            }
            --k;
            ans = ans * v % mod;
        }
        int[][] c = new int[x + 1][x + 1];
        for (int i = 0; i <= x; ++i) {
            c[i][0] = 1;
            for (int j = 1; j <= i; ++j) {
                c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
            }
        }
        ans = ((ans * c[x][k]) % mod) * qpow(val, k) % mod;
        return (int) ans;
    }

    private long qpow(long a, int n) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
func countKSubsequencesWithMaxBeauty(s string, k int) int {
	f := [26]int{}
	cnt := 0
	for _, c := range s {
		f[c-'a']++
		if f[c-'a'] == 1 {
			cnt++
		}
	}
	if cnt < k {
		return 0
	}
	vs := []int{}
	for _, x := range f {
		if x > 0 {
			vs = append(vs, x)
		}
	}
	sort.Slice(vs, func(i, j int) bool {
		return vs[i] > vs[j]
	})
	const mod int = 1e9 + 7
	ans := 1
	val := vs[k-1]
	x := 0
	for _, v := range vs {
		if v == val {
			x++
		}
	}
	for _, v := range vs {
		if v == val {
			break
		}
		k--
		ans = ans * v % mod
	}
	c := make([][]int, x+1)
	for i := range c {
		c[i] = make([]int, x+1)
		c[i][0] = 1
		for j := 1; j <= i; j++ {
			c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
		}
	}
	qpow := func(a, n int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	ans = (ans * c[x][k] % mod) * qpow(val, k) % mod
	return ans
}

# Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
class Solution:
    def countKSubsequencesWithMaxBeauty(self, s: str, k: int) -> int:
        f = Counter(s)
        if len(f) < k:
            return 0
        mod = 10**9 + 7
        vs = sorted(f.values(), reverse=True)
        val = vs[k - 1]
        x = vs.count(val)
        ans = 1
        for v in vs:
            if v == val:
                break
            k -= 1
            ans = ans * v % mod
        ans = ans * comb(x, k) * pow(val, k, mod) % mod
        return ans

// Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
// class Solution {
//     private final int mod = (int) 1e9 + 7;
// 
//     public int countKSubsequencesWithMaxBeauty(String s, int k) {
//         int[] f = new int[26];
//         int n = s.length();
//         int cnt = 0;
//         for (int i = 0; i < n; ++i) {
//             if (++f[s.charAt(i) - 'a'] == 1) {
//                 ++cnt;
//             }
//         }
//         if (cnt < k) {
//             return 0;
//         }
//         Integer[] vs = new Integer[cnt];
//         for (int i = 0, j = 0; i < 26; ++i) {
//             if (f[i] > 0) {
//                 vs[j++] = f[i];
//             }
//         }
//         Arrays.sort(vs, (a, b) -> b - a);
//         long ans = 1;
//         int val = vs[k - 1];
//         int x = 0;
//         for (int v : vs) {
//             if (v == val) {
//                 ++x;
//             }
//         }
//         for (int v : vs) {
//             if (v == val) {
//                 break;
//             }
//             --k;
//             ans = ans * v % mod;
//         }
//         int[][] c = new int[x + 1][x + 1];
//         for (int i = 0; i <= x; ++i) {
//             c[i][0] = 1;
//             for (int j = 1; j <= i; ++j) {
//                 c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
//             }
//         }
//         ans = ((ans * c[x][k]) % mod) * qpow(val, k) % mod;
//         return (int) ans;
//     }
// 
//     private long qpow(long a, int n) {
//         long ans = 1;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % mod;
//             }
//             a = a * a % mod;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2842: Count K-Subsequences of a String With Maximum Beauty
function countKSubsequencesWithMaxBeauty(s: string, k: number): number {
    const f: number[] = new Array(26).fill(0);
    let cnt = 0;
    for (const c of s) {
        const i = c.charCodeAt(0) - 97;
        if (++f[i] === 1) {
            ++cnt;
        }
    }
    if (cnt < k) {
        return 0;
    }
    const mod = BigInt(10 ** 9 + 7);
    const vs: number[] = f.filter(v => v > 0).sort((a, b) => b - a);
    const val = vs[k - 1];
    const x = vs.filter(v => v === val).length;
    let ans = 1n;
    for (const v of vs) {
        if (v === val) {
            break;
        }
        --k;
        ans = (ans * BigInt(v)) % mod;
    }
    const c: number[][] = new Array(x + 1).fill(0).map(() => new Array(k + 1).fill(0));
    for (let i = 0; i <= x; ++i) {
        c[i][0] = 1;
        for (let j = 1; j <= Math.min(i, k); ++j) {
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % Number(mod);
        }
    }
    const qpow = (a: bigint, n: number): bigint => {
        let ans = 1n;
        for (; n; n >>>= 1) {
            if (n & 1) {
                ans = (ans * a) % BigInt(mod);
            }
            a = (a * a) % BigInt(mod);
        }
        return ans;
    };
    ans = (((ans * BigInt(c[x][k])) % mod) * qpow(BigInt(val), k)) % mod;
    return Number(ans);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(|\Sigma|)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.