LeetCode #832 — EASY

Flipping an Image

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

The Problem

Problem Statement

Given an n x n binary matrix image, flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed.

For example, flipping [1,1,0] horizontally results in [0,1,1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0.

For example, inverting [0,1,1] results in [1,0,0].

Example 1:

Input: image = [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]

Example 2:

Input: image = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

Constraints:

n == image.length
n == image[i].length
1 <= n <= 20
images[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: Given an n x n binary matrix image, flip the image horizontally, then invert it, and return the resulting image. To flip an image horizontally means that each row of the image is reversed. For example, flipping [1,1,0] horizontally results in [0,1,1]. To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting [0,1,1] results in [1,0,0].

Baseline thinking

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

Pattern signal: Array · Two Pointers · Bit Manipulation

Example 1

[[1,1,0],[1,0,1],[0,0,0]]

Example 2

[[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #832: Flipping an Image
class Solution {
    public int[][] flipAndInvertImage(int[][] image) {
        for (var row : image) {
            int i = 0, j = row.length - 1;
            for (; i < j; ++i, --j) {
                if (row[i] == row[j]) {
                    row[i] ^= 1;
                    row[j] ^= 1;
                }
            }
            if (i == j) {
                row[i] ^= 1;
            }
        }
        return image;
    }
}

// Accepted solution for LeetCode #832: Flipping an Image
func flipAndInvertImage(image [][]int) [][]int {
	for _, row := range image {
		i, j := 0, len(row)-1
		for ; i < j; i, j = i+1, j-1 {
			if row[i] == row[j] {
				row[i] ^= 1
				row[j] ^= 1
			}
		}
		if i == j {
			row[i] ^= 1
		}
	}
	return image
}

# Accepted solution for LeetCode #832: Flipping an Image
class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        n = len(image)
        for row in image:
            i, j = 0, n - 1
            while i < j:
                if row[i] == row[j]:
                    row[i] ^= 1
                    row[j] ^= 1
                i, j = i + 1, j - 1
            if i == j:
                row[i] ^= 1
        return image

// Accepted solution for LeetCode #832: Flipping an Image
struct Solution;

impl Solution {
    fn flip_and_invert_image(mut a: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let n = a.len();
        for i in 0..n {
            let mut l = 0;
            let mut r = n - 1;
            while l < r {
                a[i].swap(l, r);
                l += 1;
                r -= 1;
            }
            for j in 0..n {
                a[i][j] = 1 - a[i][j];
            }
        }
        a
    }
}

#[test]
fn test() {
    let a: Vec<Vec<i32>> = vec_vec_i32![[1, 1, 0], [1, 0, 1], [0, 0, 0]];
    let b: Vec<Vec<i32>> = vec_vec_i32![[1, 0, 0], [0, 1, 0], [1, 1, 1]];
    assert_eq!(Solution::flip_and_invert_image(a), b);
    let a: Vec<Vec<i32>> = vec_vec_i32![[1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0]];
    let b: Vec<Vec<i32>> = vec_vec_i32![[1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 0, 1], [1, 0, 1, 0]];
    assert_eq!(Solution::flip_and_invert_image(a), b);
}

// Accepted solution for LeetCode #832: Flipping an Image
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #832: Flipping an Image
// class Solution {
//     public int[][] flipAndInvertImage(int[][] image) {
//         for (var row : image) {
//             int i = 0, j = row.length - 1;
//             for (; i < j; ++i, --j) {
//                 if (row[i] == row[j]) {
//                     row[i] ^= 1;
//                     row[j] ^= 1;
//                 }
//             }
//             if (i == j) {
//                 row[i] ^= 1;
//             }
//         }
//         return image;
//     }
// }

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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.