Jump Game IV.cpp

PROBLEM:



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 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. 1 <= arr.length <= 5 * 10^4
2. -10^8 <= arr[i] <= 10^8





SOLUTION:



class Solution {
public:
    int minJumps(vector<int>& arr) {
        
        int i,ans=0,n;
        map<int,vector<int>> m;
        n = arr.size();
        
        if(n==1)
            return 0;
        
        vector<bool> vis(n+1,false);
        queue<int> q;
        vis[0]=true;
        
        for(i=0;i<n;i++)
        {
            m[arr[i]].push_back(i);
        }
        

        q.push(0);
        while(!q.empty())
        {
            for (int size = q.size(); size > 0; --size) {
                int i = q.front(); q.pop();
                if (i == n - 1) return ans; // Reached to last index
                vector<int>& next = m[arr[i]];
                next.push_back(i - 1); next.push_back(i + 1);
                for (int j : next) {
                    if (j >= 0 && j < n && !vis[j]) {
                        vis[j] = true;
                        q.push(j);
                    }
                }
                next.clear(); // avoid later lookup indicesOfValue arr[i]
            }
            ans++;
        }
        
        return 0;
    }
};