<p>You are given an array of <code>words</code> where each word consists of lowercase English letters.</p>

<p><code>word<sub>A</sub></code> is a <strong>predecessor</strong> of <code>word<sub>B</sub></code> if and only if we can insert <strong>exactly one</strong> letter anywhere in <code>word<sub>A</sub></code> <strong>without changing the order of the other characters</strong> to make it equal to <code>word<sub>B</sub></code>.</p>

<ul>
	<li>For example, <code>&quot;abc&quot;</code> is a <strong>predecessor</strong> of <code>&quot;ab<u>a</u>c&quot;</code>, while <code>&quot;cba&quot;</code> is not a <strong>predecessor</strong> of <code>&quot;bcad&quot;</code>.</li>
</ul>

<p>A <strong>word chain</strong><em> </em>is a sequence of words <code>[word<sub>1</sub>, word<sub>2</sub>, ..., word<sub>k</sub>]</code> with <code>k &gt;= 1</code>, where <code>word<sub>1</sub></code> is a <strong>predecessor</strong> of <code>word<sub>2</sub></code>, <code>word<sub>2</sub></code> is a <strong>predecessor</strong> of <code>word<sub>3</sub></code>, and so on. A single word is trivially a <strong>word chain</strong> with <code>k == 1</code>.</p>

<p>Return <em>the <strong>length</strong> of the <strong>longest possible word chain</strong> with words chosen from the given list of </em><code>words</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;b&quot;,&quot;ba&quot;,&quot;bca&quot;,&quot;bda&quot;,&quot;bdca&quot;]
<strong>Output:</strong> 4
<strong>Explanation</strong>: One of the longest word chains is [&quot;a&quot;,&quot;<u>b</u>a&quot;,&quot;b<u>d</u>a&quot;,&quot;bd<u>c</u>a&quot;].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;xbc&quot;,&quot;pcxbcf&quot;,&quot;xb&quot;,&quot;cxbc&quot;,&quot;pcxbc&quot;]
<strong>Output:</strong> 5
<strong>Explanation:</strong> All the words can be put in a word chain [&quot;xb&quot;, &quot;xb<u>c</u>&quot;, &quot;<u>c</u>xbc&quot;, &quot;<u>p</u>cxbc&quot;, &quot;pcxbc<u>f</u>&quot;].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;abcd&quot;,&quot;dbqca&quot;]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The trivial word chain [&quot;abcd&quot;] is one of the longest word chains.
[&quot;abcd&quot;,&quot;dbqca&quot;] is not a valid word chain because the ordering of the letters is changed.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 16</code></li>
	<li><code>words[i]</code> only consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    // Tells whether prevWord is predecessor of lastWord or not 
    bool isPredecessor(string &currWord, string &prevWord) {
        // Current word be 1 greater in size than prev word
        if(currWord.size() != prevWord.size() + 1) return false;

        int i=0, j=0, cntChanges = 0;
        while(i < prevWord.size()) {
            if(prevWord[i] == currWord[j]) {
                i++, j++;
            } else {
                cntChanges++;
                j++;
            }

            // If more than 1 changes, means ordered is disturbed (1 changes is acceptable bcoz of new character present in currWord)
            if(cntChanges > 1) return false;
        }

        return true;
    }

    int longestStrChain(vector<string>& words) {
        int n = words.size();

        // Sort according to size of strings
        sort(words.begin(), words.end(), [](const std::string &a, const std::string &b) { 
            return a.size() < b.size();
        });

        vector<vector<int>> dp(n+1, vector<int> (n+1, -1));    // {ind, prevInd}
        vector<int> nextRow(n+1, -1);    // {prevInd}
        // Since, prevInd start from -1, so use Co-ordinate change & map prevInd values 1 index ahead 

        // Base case: Exhausted all the oindexes, so no further elements remaining to be part of answer
        for(int prevInd=-1; prevInd<n; prevInd++) nextRow[prevInd + 1] = 0;

        for(int ind=n-1; ind>=0; ind--) {

            vector<int> currRow(n+1);
            for(int prevInd=ind-1; prevInd>=-1; prevInd--) {
                int pick = ((prevInd == -1) || (isPredecessor(words[ind], words[prevInd]))) ? 1 + nextRow[ind + 1] : 0;
                int notPick = 0 + nextRow[prevInd + 1];

                currRow[prevInd + 1] = max(pick, notPick);
            }

            nextRow = currRow;
        }

        return nextRow[-1 + 1];    // Tells maxLen of LSC starting from 0th index with no previous elements
    }
};
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