LeetCode #3071 — MEDIUM

Minimum Operations to Write the Letter Y on a Grid

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

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed n x n grid where n is odd, and grid[r][c] is 0, 1, or 2.

We say that a cell belongs to the Letter Y if it belongs to one of the following:

The diagonal starting at the top-left cell and ending at the center cell of the grid.
The diagonal starting at the top-right cell and ending at the center cell of the grid.
The vertical line starting at the center cell and ending at the bottom border of the grid.

The Letter Y is written on the grid if and only if:

All values at cells belonging to the Y are equal.
All values at cells not belonging to the Y are equal.
The values at cells belonging to the Y are different from the values at cells not belonging to the Y.

Return the minimum number of operations needed to write the letter Y on the grid given that in one operation you can change the value at any cell to 0, 1, or 2.

Example 1:

Input: grid = [[1,2,2],[1,1,0],[0,1,0]]
Output: 3
Explanation: We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 1 while those that do not belong to Y are equal to 0.
It can be shown that 3 is the minimum number of operations needed to write Y on the grid.

Example 2:

Input: grid = [[0,1,0,1,0],[2,1,0,1,2],[2,2,2,0,1],[2,2,2,2,2],[2,1,2,2,2]]
Output: 12
Explanation: We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 0 while those that do not belong to Y are equal to 2. 
It can be shown that 12 is the minimum number of operations needed to write Y on the grid.

Constraints:

3 <= n <= 49
n == grid.length == grid[i].length
0 <= grid[i][j] <= 2
n is odd.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed n x n grid where n is odd, and grid[r][c] is 0, 1, or 2. We say that a cell belongs to the Letter Y if it belongs to one of the following: The diagonal starting at the top-left cell and ending at the center cell of the grid. The diagonal starting at the top-right cell and ending at the center cell of the grid. The vertical line starting at the center cell and ending at the bottom border of the grid. The Letter Y is written on the grid if and only if: All values at cells belonging to the Y are equal. All values at cells not belonging to the Y are equal. The values at cells belonging to the Y are different from the values at cells not belonging to the Y. Return the minimum number of operations needed to write the letter Y on the grid given that in one operation you can change the value at any cell to 0, 1, or 2.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[[1,2,2],[1,1,0],[0,1,0]]

Example 2

[[0,1,0,1,0],[2,1,0,1,2],[2,2,2,0,1],[2,2,2,2,2],[2,1,2,2,2]]

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 #3071: Minimum Operations to Write the Letter Y on a Grid
class Solution {
    public int minimumOperationsToWriteY(int[][] grid) {
        int n = grid.length;
        int[] cnt1 = new int[3];
        int[] cnt2 = new int[3];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                boolean a = i == j && i <= n / 2;
                boolean b = i + j == n - 1 && i <= n / 2;
                boolean c = j == n / 2 && i >= n / 2;
                if (a || b || c) {
                    ++cnt1[grid[i][j]];
                } else {
                    ++cnt2[grid[i][j]];
                }
            }
        }
        int ans = n * n;
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (i != j) {
                    ans = Math.min(ans, n * n - cnt1[i] - cnt2[j]);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3071: Minimum Operations to Write the Letter Y on a Grid
func minimumOperationsToWriteY(grid [][]int) int {
	n := len(grid)
	cnt1 := [3]int{}
	cnt2 := [3]int{}
	for i, row := range grid {
		for j, x := range row {
			a := i == j && i <= n/2
			b := i+j == n-1 && i <= n/2
			c := j == n/2 && i >= n/2
			if a || b || c {
				cnt1[x]++
			} else {
				cnt2[x]++
			}
		}
	}
	ans := n * n
	for i := 0; i < 3; i++ {
		for j := 0; j < 3; j++ {
			if i != j {
				ans = min(ans, n*n-cnt1[i]-cnt2[j])
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #3071: Minimum Operations to Write the Letter Y on a Grid
class Solution:
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:
        n = len(grid)
        cnt1 = Counter()
        cnt2 = Counter()
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                a = i == j and i <= n // 2
                b = i + j == n - 1 and i <= n // 2
                c = j == n // 2 and i >= n // 2
                if a or b or c:
                    cnt1[x] += 1
                else:
                    cnt2[x] += 1
        return min(
            n * n - cnt1[i] - cnt2[j] for i in range(3) for j in range(3) if i != j
        )

// Accepted solution for LeetCode #3071: Minimum Operations to Write the Letter Y on a Grid
// 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 #3071: Minimum Operations to Write the Letter Y on a Grid
// class Solution {
//     public int minimumOperationsToWriteY(int[][] grid) {
//         int n = grid.length;
//         int[] cnt1 = new int[3];
//         int[] cnt2 = new int[3];
//         for (int i = 0; i < n; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 boolean a = i == j && i <= n / 2;
//                 boolean b = i + j == n - 1 && i <= n / 2;
//                 boolean c = j == n / 2 && i >= n / 2;
//                 if (a || b || c) {
//                     ++cnt1[grid[i][j]];
//                 } else {
//                     ++cnt2[grid[i][j]];
//                 }
//             }
//         }
//         int ans = n * n;
//         for (int i = 0; i < 3; ++i) {
//             for (int j = 0; j < 3; ++j) {
//                 if (i != j) {
//                     ans = Math.min(ans, n * n - cnt1[i] - cnt2[j]);
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3071: Minimum Operations to Write the Letter Y on a Grid
function minimumOperationsToWriteY(grid: number[][]): number {
    const n = grid.length;
    const cnt1: number[] = Array(3).fill(0);
    const cnt2: number[] = Array(3).fill(0);
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            const a = i === j && i <= n >> 1;
            const b = i + j === n - 1 && i <= n >> 1;
            const c = j === n >> 1 && i >= n >> 1;
            if (a || b || c) {
                ++cnt1[grid[i][j]];
            } else {
                ++cnt2[grid[i][j]];
            }
        }
    }
    let ans = n * n;
    for (let i = 0; i < 3; ++i) {
        for (let j = 0; j < 3; ++j) {
            if (i !== j) {
                ans = Math.min(ans, n * n - cnt1[i] - cnt2[j]);
            }
        }
    }
    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

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.