Greedy/ C++.
IntuitionThe problem at hand involves finding a sequence of indices in word1 such that the sequence of characters from word1 corresponds to the subsequence word
# Intuition
The problem at hand involves finding a sequence of indices in word1 such that the sequence of characters from word1 corresponds to the subsequence word2. The challenge is to find the subsequence of word2 in word1 such that the indices form a valid sequence. To achieve this, we must ensure that the order of characters in word2 is maintained while scanning through word1. Additionally, we are allowed to skip characters in word1 to match word2 and must account for the last occurrence of characters in word2 within word1.

---

# Approach
The approach begins by processing the characters of word1 from right to left, and the characters of word2 from right to left, and storing the last index of each character of word2 found in word1. This is done using the locc array, which tracks the index in word1 where each character of word2 last appears.

Next, the algorithm processes word1 from left to right, trying to form the subsequence for word2. It compares each character of word1 to the current character of word2. If there's a match, it adds the index to the result list and proceeds to the next character of word2. If the current character doesn't match but is still required to form a valid subsequence, the convert flag is set, allowing it to skip characters and still match the remaining characters of word2.

Finally, if the subsequence matches the length of word2, the result is returned; otherwise, an empty vector is returned, indicating no valid subsequence was found. This approach efficiently ensures the correct sequence of indices is found while minimizing unnecessary checks.

---

# Complexity
- Time complexity:
O(n+m)

- Space complexity:
O(m)

# Code
```cpp []
class Solution {
public:
    vector<int> validSequence(string word1, string word2) {
        int n = word1.size(),m = word2.size();
        vector<int> locc(m,-1),ans;
        for(int i=n-1,j=m-1;i>=0;i--)
        {
            if(j>=0 && word1[i]==word2[j])
            locc[j--] = i;
        }
        int convert=0;
        for(int i=0,j=0;i<n && j<m;i++)
        {
            if(word1[i]==word2[j] || (!convert && (j==m-1 || locc[j+1]>i)))
            {
                ans.push_back(i);
                if(word1[i] != word2[j])
                convert = 1;
                j++;
            }
        }
        return ans.size()==m?ans:vector<int>();
    }
};
```
['C++']