LeetCode #1897 — EASY

Redistribute Characters to Make All Strings Equal

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

You are given an array of strings words (0-indexed).

In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j].

Return true if you can make every string in words equal using any number of operations, and false otherwise.

Example 1:

Input: words = ["abc","aabc","bc"]
Output: true
Explanation: Move the first 'a' in words[1] to the front of words[2],
to make words[1] = "abc" and words[2] = "abc".
All the strings are now equal to "abc", so return true.

Example 2:

Input: words = ["ab","a"]
Output: false
Explanation: It is impossible to make all the strings equal using the operation.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings words (0-indexed). In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j]. Return true if you can make every string in words equal using any number of operations, and false otherwise.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

["abc","aabc","bc"]

Example 2

["ab","a"]

Step 02

Core Insight

What unlocks the optimal approach

Characters are independent—only the frequency of characters matters.
It is possible to distribute characters if all characters can be divided equally among all strings.

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 #1897: Redistribute Characters to Make All Strings Equal
class Solution {
    public boolean makeEqual(String[] words) {
        int[] cnt = new int[26];
        for (var w : words) {
            for (char c : w.toCharArray()) {
                ++cnt[c - 'a'];
            }
        }
        int n = words.length;
        for (int v : cnt) {
            if (v % n != 0) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1897: Redistribute Characters to Make All Strings Equal
func makeEqual(words []string) bool {
	cnt := [26]int{}
	for _, w := range words {
		for _, c := range w {
			cnt[c-'a']++
		}
	}
	n := len(words)
	for _, v := range cnt {
		if v%n != 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #1897: Redistribute Characters to Make All Strings Equal
class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        cnt = Counter()
        for w in words:
            for c in w:
                cnt[c] += 1
        n = len(words)
        return all(v % n == 0 for v in cnt.values())

// Accepted solution for LeetCode #1897: Redistribute Characters to Make All Strings Equal
impl Solution {
    pub fn make_equal(words: Vec<String>) -> bool {
        let mut cnt = std::collections::HashMap::new();

        for word in words.iter() {
            for c in word.chars() {
                *cnt.entry(c).or_insert(0) += 1;
            }
        }

        let n = words.len();
        cnt.values().all(|&v| v % n == 0)
    }
}

// Accepted solution for LeetCode #1897: Redistribute Characters to Make All Strings Equal
function makeEqual(words: string[]): boolean {
    const cnt: Record<string, number> = {};
    for (const w of words) {
        for (const c of w) {
            cnt[c] = (cnt[c] || 0) + 1;
        }
    }
    const n = words.length;
    return Object.values(cnt).every(v => v % n === 0);
}

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(L)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

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.