Reverse String

EASY✓ CHIP-TIMEDLC #344 — FULL STATEMENT

The briefingthe full drill, in my own words

The drill: Flip a character array end-to-end by mutating the buffer you were handed — no second array. Encoded here as an array of one-character strings, exactly how LeetCode's Python signature types it.

A buffer of single characters arrives, and the task is to flip it end-to-end so the last character becomes first and the first becomes last, with nothing returned separately.

The array itself is the answer: every character keeps its identity, only positions change, and the reversal has to happen inside the same buffer rather than producing a fresh one.

Whitespace and repeated characters are ordinary members of the array with no special handling; an array of length one or zero simply reverses to itself.

buffer holds anywhere from zero to a few thousand characters
characters may repeat and include spaces or punctuation
the buffer must be edited in place — no second array returned
extra space beyond a couple of index variables is off budget

EX 01

s = ["g", "r", "i", "n", "d"]

→ ["d", "n", "i", "r", "g"]

EX 02

s = ["a"]

→ ["a"]

single element — nothing to swap

EX 03

s = ["x", "y"]

→ ["y", "x"]

smallest real swap

The hintstake only what you need

HINT 1 THE NUDGE

Building the reversed copy and pasting it back is correct — and it allocates precisely the buffer the follow-up dares you to live without.

HINT 2 THE STRUCTURE

Reversal is nothing but independent swaps: position i trades places with position n−1−i. No intermediate storage is ever required for that trade.

HINT 3 ONE STEP FROM THE ANSWER

Two pointers, one at each end: swap what they hold, step both inward, stop when they meet. The middle element of an odd-length array never has to move.

Coach's boardthe pattern, step by step

SWAP FROM BOTH ENDSPATTERN · TWO POINTERSs = ["g", "r", "i", "n", "d"]

STEP 1

Two pointers, L at 0 and R at 4 — swap what they hold and step inward.

STEP 1 / 6 · ← → WORK TOO

The splitstwo paces, two languages

grind/reverse-string.py — RACE PACE

LANG ▸

PACE ▸

class Solution:
    def reverseString(self, s: List[str]) -> None:
        l, r = 0, len(s) - 1
        while l < r:
            s[l], s[r] = s[r], s[l]
            l += 1
            r -= 1

TIME O(N)SPACE O(1)PYTHON · RACE PACE · 7 LN

class Solution:
    def reverseString(self, s: List[str]) -> None:
        rev = s[::-1]  # the extra buffer the follow-up forbids
        for i in range(len(s)):
            s[i] = rev[i]

TIME O(N)SPACE O(N)PYTHON · TRAINING PACE · 5 LN

class Solution {
    public void reverseString(String[] s) {
        int l = 0, r = s.length - 1;
        while (l < r) {
            String tmp = s[l];
            s[l] = s[r];
            s[r] = tmp;
            l++;
            r--;
        }
    }
}

TIME O(N)SPACE O(1)JAVA · RACE PACE · 12 LN

class Solution {
    public void reverseString(String[] s) {
        String[] rev = new String[s.length]; // the extra buffer the follow-up forbids
        for (int i = 0; i < s.length; i++) {
            rev[i] = s[s.length - 1 - i];
        }
        for (int i = 0; i < s.length; i++) {
            s[i] = rev[i];
        }
    }
}

TIME O(N)SPACE O(N)JAVA · TRAINING PACE · 11 LN

✓ CHIP-TIMEDALL 4 SOLUTIONS RAN GREEN AGAINST SELF-AUTHORED CASES IN CI · JDK 21 · CPYTHON 3.12 · NOTHING PUBLISHES RED