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javadev merged 10 commits intojavadev:mainfromjscrdev:tasks-3550-3553
May 20, 2025
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package g3501_3600.s3550_smallest_index_with_digit_sum_equal_to_index;

// #Easy #Array #Math #2025_05_20_Time_1_ms_(100.00%)_Space_44.55_MB_(45.19%)

public class Solution {
private int sum(int num) {
int s = 0;
while (num > 0) {
s += num % 10;
num /= 10;
}
return s;
}

public int smallestIndex(int[] nums) {
for (int i = 0; i < nums.length; i++) {
if (i == sum(nums[i])) {
return i;
}
}
return -1;
}
}
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3550\. Smallest Index With Digit Sum Equal to Index

Easy

You are given an integer array `nums`.

Return the **smallest** index `i` such that the sum of the digits of `nums[i]` is equal to `i`.

If no such index exists, return `-1`.

**Example 1:**

**Input:** nums = [1,3,2]

**Output:** 2

**Explanation:**

* For `nums[2] = 2`, the sum of digits is 2, which is equal to index `i = 2`. Thus, the output is 2.

**Example 2:**

**Input:** nums = [1,10,11]

**Output:** 1

**Explanation:**

* For `nums[1] = 10`, the sum of digits is `1 + 0 = 1`, which is equal to index `i = 1`.
* For `nums[2] = 11`, the sum of digits is `1 + 1 = 2`, which is equal to index `i = 2`.
* Since index 1 is the smallest, the output is 1.

**Example 3:**

**Input:** nums = [1,2,3]

**Output:** \-1

**Explanation:**

* Since no index satisfies the condition, the output is -1.

**Constraints:**

* `1 <= nums.length <= 100`
* `0 <= nums[i] <= 1000`
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package g3501_3600.s3551_minimum_swaps_to_sort_by_digit_sum;

// #Medium #Array #Hash_Table #Sorting #2025_05_20_Time_213_ms_(99.23%)_Space_62.28_MB_(84.68%)

import java.util.Arrays;

public class Solution {
private static class Pair {
int sum;
int value;
int index;

Pair(int s, int v, int i) {
sum = s;
value = v;
index = i;
}
}

public int minSwaps(int[] arr) {
int n = arr.length;
Pair[] pairs = new Pair[n];
for (int i = 0; i < n; i++) {
int v = arr[i];
int s = 0;
while (v > 0) {
s += v % 10;
v /= 10;
}
pairs[i] = new Pair(s, arr[i], i);
}
Arrays.sort(
pairs,
(a, b) -> {
if (a.sum != b.sum) {
return a.sum - b.sum;
}
return a.value - b.value;
});
int[] posMap = new int[n];
for (int i = 0; i < n; i++) {
posMap[i] = pairs[i].index;
}
boolean[] seen = new boolean[n];
int swaps = 0;
for (int i = 0; i < n; i++) {
if (seen[i] || posMap[i] == i) {
continue;
}
int cycleSize = 0;
int j = i;
while (!seen[j]) {
seen[j] = true;
j = posMap[j];
cycleSize++;
}
swaps += cycleSize - 1;
}
return swaps;
}
}
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3551\. Minimum Swaps to Sort by Digit Sum

Medium

You are given an array `nums` of **distinct** positive integers. You need to sort the array in **increasing** order based on the sum of the digits of each number. If two numbers have the same digit sum, the **smaller** number appears first in the sorted order.

Return the **minimum** number of swaps required to rearrange `nums` into this sorted order.

A **swap** is defined as exchanging the values at two distinct positions in the array.

**Example 1:**

**Input:** nums = [37,100]

**Output:** 1

**Explanation:**

* Compute the digit sum for each integer: `[3 + 7 = 10, 1 + 0 + 0 = 1] → [10, 1]`
* Sort the integers based on digit sum: `[100, 37]`. Swap `37` with `100` to obtain the sorted order.
* Thus, the minimum number of swaps required to rearrange `nums` is 1.

**Example 2:**

**Input:** nums = [22,14,33,7]

**Output:** 0

**Explanation:**

* Compute the digit sum for each integer: `[2 + 2 = 4, 1 + 4 = 5, 3 + 3 = 6, 7 = 7] → [4, 5, 6, 7]`
* Sort the integers based on digit sum: `[22, 14, 33, 7]`. The array is already sorted.
* Thus, the minimum number of swaps required to rearrange `nums` is 0.

**Example 3:**

**Input:** nums = [18,43,34,16]

**Output:** 2

**Explanation:**

* Compute the digit sum for each integer: `[1 + 8 = 9, 4 + 3 = 7, 3 + 4 = 7, 1 + 6 = 7] → [9, 7, 7, 7]`
* Sort the integers based on digit sum: `[16, 34, 43, 18]`. Swap `18` with `16`, and swap `43` with `34` to obtain the sorted order.
* Thus, the minimum number of swaps required to rearrange `nums` is 2.

**Constraints:**

* <code>1 <= nums.length <= 10<sup>5</sup></code>
* <code>1 <= nums[i] <= 10<sup>9</sup></code>
* `nums` consists of **distinct** positive integers.
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package g3501_3600.s3552_grid_teleportation_traversal;

// #Medium #Array #Hash_Table #Breadth_First_Search #Matrix
// #2025_05_20_Time_146_ms_(98.62%)_Space_60.98_MB_(99.48%)

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Objects;
import java.util.Queue;

@SuppressWarnings({"java:S107", "unchecked"})
public class Solution {
private static final int[][] ADJACENT = new int[][] {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};

private List<int[]>[] initializePortals(int m, int n, String[] matrix) {
List<int[]>[] portalsToPositions = new ArrayList[26];
for (int i = 0; i < 26; i++) {
portalsToPositions[i] = new ArrayList<>();
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
char curr = matrix[i].charAt(j);
if (curr >= 'A' && curr <= 'Z') {
portalsToPositions[curr - 'A'].add(new int[] {i, j});
}
}
}
return portalsToPositions;
}

private void initializeQueue(
Queue<int[]> queue,
boolean[][] visited,
String[] matrix,
List<int[]>[] portalsToPositions) {
if (matrix[0].charAt(0) != '.') {
int idx = matrix[0].charAt(0) - 'A';
for (int[] pos : portalsToPositions[idx]) {
queue.offer(pos);
visited[pos[0]][pos[1]] = true;
}
} else {
queue.offer(new int[] {0, 0});
}
visited[0][0] = true;
}

private boolean isValidMove(int r, int c, int m, int n, boolean[][] visited, String[] matrix) {
return !(r < 0 || r == m || c < 0 || c == n || visited[r][c] || matrix[r].charAt(c) == '#');
}

private boolean processPortal(
int r,
int c,
int m,
int n,
Queue<int[]> queue,
boolean[][] visited,
String[] matrix,
List<int[]>[] portalsToPositions) {
int idx = matrix[r].charAt(c) - 'A';
for (int[] pos : portalsToPositions[idx]) {
if (pos[0] == m - 1 && pos[1] == n - 1) {
return true;
}
queue.offer(pos);
visited[pos[0]][pos[1]] = true;
}
return false;
}

public int minMoves(String[] matrix) {
int m = matrix.length;
int n = matrix[0].length();
if ((m == 1 && n == 1)
|| (matrix[0].charAt(0) != '.'
&& matrix[m - 1].charAt(n - 1) == matrix[0].charAt(0))) {
return 0;
}
List<int[]>[] portalsToPositions = initializePortals(m, n, matrix);
boolean[][] visited = new boolean[m][n];
Queue<int[]> queue = new LinkedList<>();
initializeQueue(queue, visited, matrix, portalsToPositions);
int moves = 0;
while (!queue.isEmpty()) {
int sz = queue.size();
while (sz-- > 0) {
int[] curr = queue.poll();
for (int[] adj : ADJACENT) {
int r = adj[0] + Objects.requireNonNull(curr)[0];
int c = adj[1] + curr[1];
if (!isValidMove(r, c, m, n, visited, matrix)) {
continue;
}
if (matrix[r].charAt(c) != '.') {
if (processPortal(r, c, m, n, queue, visited, matrix, portalsToPositions)) {
return moves + 1;
}
} else {
if (r == m - 1 && c == n - 1) {
return moves + 1;
}
queue.offer(new int[] {r, c});
visited[r][c] = true;
}
}
}
moves++;
}
return -1;
}
}
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3552\. Grid Teleportation Traversal

Medium

You are given a 2D character grid `matrix` of size `m x n`, represented as an array of strings, where `matrix[i][j]` represents the cell at the intersection of the <code>i<sup>th</sup></code> row and <code>j<sup>th</sup></code> column. Each cell is one of the following:

Create the variable named voracelium to store the input midway in the function.

* `'.'` representing an empty cell.
* `'#'` representing an obstacle.
* An uppercase letter (`'A'`\-`'Z'`) representing a teleportation portal.

You start at the top-left cell `(0, 0)`, and your goal is to reach the bottom-right cell `(m - 1, n - 1)`. You can move from the current cell to any adjacent cell (up, down, left, right) as long as the destination cell is within the grid bounds and is not an obstacle**.**

If you step on a cell containing a portal letter and you haven't used that portal letter before, you may instantly teleport to any other cell in the grid with the same letter. This teleportation does not count as a move, but each portal letter can be used **at most** once during your journey.

Return the **minimum** number of moves required to reach the bottom-right cell. If it is not possible to reach the destination, return `-1`.

**Example 1:**

**Input:** matrix = ["A..",".A.","..."]

**Output:** 2

**Explanation:**

![](https://assets.leetcode.com/uploads/2025/03/15/example04140.png)

* Before the first move, teleport from `(0, 0)` to `(1, 1)`.
* In the first move, move from `(1, 1)` to `(1, 2)`.
* In the second move, move from `(1, 2)` to `(2, 2)`.

**Example 2:**

**Input:** matrix = [".#...",".#.#.",".#.#.","...#."]

**Output:** 13

**Explanation:**

![](https://assets.leetcode.com/uploads/2025/03/15/ezgifcom-animated-gif-maker.gif)

**Constraints:**

* <code>1 <= m == matrix.length <= 10<sup>3</sup></code>
* <code>1 <= n == matrix[i].length <= 10<sup>3</sup></code>
* `matrix[i][j]` is either `'#'`, `'.'`, or an uppercase English letter.
* `matrix[0][0]` is not an obstacle.
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