# 1345. Jump Game IV

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 = [7]
Output: 0
Explanation: Start index is the last index. You do not 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.

Constraints:

• `1 <= arr.length <= 5 * 104`
• `-108 <= arr[i] <= 108`

``````class Solution {
public:
int minJumps(vector<int>& arr) {
int res = 0, n = arr.size();
unordered_map<int, vector<int>> m;
queue<int> q{{0}};
for (int i = 0; i < n; ++i) {
m[arr[i]].push_back(i);
}
while (!q.empty()) {
for (int i = q.size(); i > 0; --i) {
int t = q.front(); q.pop();
if (t == n - 1) return res;
if (t > 0 && m.count(arr[t - 1])) {
q.push(t - 1);
}
if (t < n - 1 && m.count(arr[t + 1])) {
q.push(t + 1);
}
if (m.count(arr[t])) {
for (int idx : m[arr[t]]) {
if (idx != t) {
q.push(idx);
}
}
}
m.erase(arr[t]);
}
++res;
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1345

Jump Game

Jump Game II

Jump Game III

Jump Game V

Jump Game VI

Jump Game VII

Jump Game VIII

https://leetcode.com/problems/jump-game-iv/

https://leetcode.com/problems/jump-game-iv/solutions/502699/java-c-bfs-solution-clean-code-o-n/

https://leetcode.com/problems/jump-game-iv/solutions/1690813/best-explanation-ever-possible-for-this-question/

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