LeetCode #1647 — MEDIUM

Minimum Deletions to Make Character Frequencies Unique

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

A string s is called good if there are no two different characters in s that have the same frequency.

Given a string s, return the minimum number of characters you need to delete to make s good.

The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab", the frequency of 'a' is 2, while the frequency of 'b' is 1.

Example 1:

Input: s = "aab"
Output: 0
Explanation: s is already good.

Example 2:

Input: s = "aaabbbcc"
Output: 2
Explanation: You can delete two 'b's resulting in the good string "aaabcc".
Another way it to delete one 'b' and one 'c' resulting in the good string "aaabbc".

Example 3:

Input: s = "ceabaacb"
Output: 2
Explanation: You can delete both 'c's resulting in the good string "eabaab".
Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).

Constraints:

1 <= s.length <= 10⁵
s contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: A string s is called good if there are no two different characters in s that have the same frequency. Given a string s, return the minimum number of characters you need to delete to make s good. The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab", the frequency of 'a' is 2, while the frequency of 'b' is 1.

Baseline thinking

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

Pattern signal: Hash Map · Greedy

Example 1

"aab"

Example 2

"aaabbbcc"

Example 3

"ceabaacb"

Core Insight

What unlocks the optimal approach

As we can only delete characters, if we have multiple characters having the same frequency, we must decrease all the frequencies of them, except one.
Sort the alphabet characters by their frequencies non-increasingly.
Iterate on the alphabet characters, keep decreasing the frequency of the current character until it reaches a value that has not appeared before.

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

Upper-end input sizes

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 #1647: Minimum Deletions to Make Character Frequencies Unique
class Solution {
    public int minDeletions(String s) {
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        Arrays.sort(cnt);
        int ans = 0;
        for (int i = 24; i >= 0; --i) {
            while (cnt[i] >= cnt[i + 1] && cnt[i] > 0) {
                --cnt[i];
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1647: Minimum Deletions to Make Character Frequencies Unique
func minDeletions(s string) (ans int) {
	cnt := make([]int, 26)
	for _, c := range s {
		cnt[c-'a']++
	}
	sort.Sort(sort.Reverse(sort.IntSlice(cnt)))
	for i := 1; i < 26; i++ {
		for cnt[i] >= cnt[i-1] && cnt[i] > 0 {
			cnt[i]--
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1647: Minimum Deletions to Make Character Frequencies Unique
class Solution:
    def minDeletions(self, s: str) -> int:
        cnt = Counter(s)
        ans, pre = 0, inf
        for v in sorted(cnt.values(), reverse=True):
            if pre == 0:
                ans += v
            elif v >= pre:
                ans += v - pre + 1
                pre -= 1
            else:
                pre = v
        return ans

// Accepted solution for LeetCode #1647: Minimum Deletions to Make Character Frequencies Unique
impl Solution {
    #[allow(dead_code)]
    pub fn min_deletions(s: String) -> i32 {
        let mut cnt = vec![0; 26];
        let mut ans = 0;

        for c in s.chars() {
            cnt[((c as u8) - ('a' as u8)) as usize] += 1;
        }

        cnt.sort_by(|&lhs, &rhs| rhs.cmp(&lhs));

        for i in 1..26 {
            while cnt[i] >= cnt[i - 1] && cnt[i] > 0 {
                cnt[i] -= 1;
                ans += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #1647: Minimum Deletions to Make Character Frequencies Unique
function minDeletions(s: string): number {
    let map = {};
    for (let c of s) {
        map[c] = (map[c] || 0) + 1;
    }
    let ans = 0;
    let vals: number[] = Object.values(map);
    vals.sort((a, b) => a - b);
    for (let i = 1; i < vals.length; ++i) {
        while (vals[i] > 0 && i != vals.indexOf(vals[i])) {
            --vals[i];
            ++ans;
        }
    }
    return 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 + |\Sigma| × log |\Sigma|)

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.

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.