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# 1345.jump-game-iv

## Problem Description

Given an array of integers arr, you are initially positioned at the first index of the array.
In one step you can jump from index i to index:
- i + 1 where: i + 1 < arr.length.
- i - 1 where: i - 1 >= 0.
- j where: arr[i] == arr[j] and i != j.
Return the minimum number of steps to reach the last index of the array.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.
Example 2:
Input: arr = 
Output: 0
Explanation: Start index is the last index. You don't need to jump.
Example 3:
Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.
Example 4:
Input: arr = [6,1,9]
Output: 2
Example 5:
Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3
Constraints:
- 1 <= arr.length <= 5 * 10^4
- -10^8 <= arr[i] <= 10^8

## Solution

This problem can use BFS to check, check each level, if already found in last index, then return, after each level, jumps+1. here level means 3 possible options to jump. so that we can find the min jump steps.
ways to jump from index `i`: 1. i + 1, if i + 1 < len && `i + 1` index not visited before 2. `i - 1`, if `i - 1 >= 0` && `i - 1` index not visited before. 3. different indexes which has the same value as value in index `i`.
Idea is to using Map to store element and corresponding indexes as key, value pair.
Queue to store each level indexes can jumped into from index `i`:
• first check curr index `i == len - 1 (last index)`
• if yes, terminate process, return jumps.
• if not, continue
• add `i + 1` if meet requirement, and mark as visited.
• add `i - 1` if meet requirement, and mark as visited
• add `map.get(arr[i])` -- the list of indexes if meet requirement, and mark visited. since it is already visited all indexes, then remove current element from map to avoid unnecessary visit.
• after each level, jumps+1
For example: Jump Game IV

### Complexity Analysis

• Time Complexity: `O(n)`
• Space Complexity: `O(n)`
`n - n is length of arr`

## Key Points

• Group each size k subarray.
• Calculate avg of subarray of size k, compare with threshold
• Count increase 1 if meet requirements

## Code

Java Code
class Solution {
public int minJumps(int[] arr) {
if (arr.length == 1) return 0;
int jumps = 0;
int len = arr.length;
// add all indexes for the same value into map, key as element, value is the list of indexes for the element
Map<Integer, List<Integer>> map = new HashMap<>();
for (int i = 0; i < len; i++) {
}
// queue to keep index
// start from index 0
queue.offer(0);
// using HashSet to store already visited index, avoid dup visit
Set<Integer> visited = new HashSet<>();
while (!queue.isEmpty()) {
int size = queue.size();
while (size-- > 0) {
int currIdx = queue.poll();
if (currIdx = len - 1) return jumps;
// check 3 options to jump
// 1) i + 1, if i + 1 not visited, add into visited, and add into queue
if (currIdx + 1 < len && visited.add(currIdx + 1)) {
queue.offer(currIdx + 1);
}
// 2) i - 1, if i - 1 not visited, add into visited and queue
if (currIdx - 1 >= 0 && visited.add(currIdx - 1)) {
queue.offer(currIdx - 1);
}
// 3) the same value with different indexes
if (map.containsKey(arr[currIdx])) {
for (int idx : map.get(arr[currIdx])) {