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Merge pull request #588 from Abhishek-Mallick/leetcode4
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‎1048-longest-string-chain/1048-longest-string-chain.java

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class Solution {
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public int longestStrChain(String[] words) {
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Map<String,Integer> map = new HashMap<>();
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Arrays.sort(words, (a,b)->a.length() - b.length());
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int res = 0;
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for(String word : words)
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{
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int best = 0;
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for(int i=0;i<word.length();i++)
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{
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String prev = word.substring(0,i) + word.substring(i+1);
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best = Math.max(best,map.getOrDefault(prev,0)+1);
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}
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map.put(word,best);
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res = Math.max(res,best);
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}
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return res;
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}
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}

‎1048-longest-string-chain/README.md

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<h2><a href="https://leetcode.com/problems/longest-string-chain/">1048. Longest String Chain</a></h2><h3>Medium</h3><hr><div><p>You are given an array of <code>words</code> where each word consists of lowercase English letters.</p>
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<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>
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<ul>
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<li>For example, <code>"abc"</code> is a <strong>predecessor</strong> of <code>"ab<u>a</u>c"</code>, while <code>"cba"</code> is not a <strong>predecessor</strong> of <code>"bcad"</code>.</li>
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</ul>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre><strong>Input:</strong> words = ["a","b","ba","bca","bda","bdca"]
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<strong>Output:</strong> 4
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<strong>Explanation</strong>: One of the longest word chains is ["a","<u>b</u>a","b<u>d</u>a","bd<u>c</u>a"].
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> All the words can be put in a word chain ["xb", "xb<u>c</u>", "<u>c</u>xbc", "<u>p</u>cxbc", "pcxbc<u>f</u>"].
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre><strong>Input:</strong> words = ["abcd","dbqca"]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> The trivial word chain ["abcd"] is one of the longest word chains.
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["abcd","dbqca"] is not a valid word chain because the ordering of the letters is changed.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= words.length &lt;= 1000</code></li>
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<li><code>1 &lt;= words[i].length &lt;= 16</code></li>
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<li><code>words[i]</code> only consists of lowercase English letters.</li>
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</ul>
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</div>

‎1480-running-sum-of-1d-array/1480-running-sum-of-1d-array.cpp

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class Solution {
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public:
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vector<int> runningSum(vector<int>& nums) {
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for(int i=1;i<nums.size();i++)
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nums[i]+= nums[i-1];
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return nums;
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}
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};

‎1480-running-sum-of-1d-array/1480-running-sum-of-1d-array.java

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class Solution {
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public int[] runningSum(int[] nums) {
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for(int i=1;i<nums.length;i++)
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nums[i]+=nums[i-1];
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return nums;
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}
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}

‎1480-running-sum-of-1d-array/NOTES.md

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​

‎1480-running-sum-of-1d-array/README.md

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<h2><a href="https://leetcode.com/problems/running-sum-of-1d-array/">1480. Running Sum of 1d Array</a></h2><h3>Easy</h3><hr><div><p>Given an array <code>nums</code>. We define a running sum of an array as&nbsp;<code>runningSum[i] = sum(nums[0]…nums[i])</code>.</p>
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<p>Return the running sum of <code>nums</code>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre><strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> [1,3,6,10]
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<strong>Explanation:</strong> Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].</pre>
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<p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> nums = [1,1,1,1,1]
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<strong>Output:</strong> [1,2,3,4,5]
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<strong>Explanation:</strong> Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].</pre>
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<p><strong>Example 3:</strong></p>
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<pre><strong>Input:</strong> nums = [3,1,2,10,1]
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<strong>Output:</strong> [3,4,6,16,17]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
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<li><code>-10^6&nbsp;&lt;= nums[i] &lt;=&nbsp;10^6</code></li>
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</ul></div>

‎1642-furthest-building-you-can-reach/1642-furthest-building-you-can-reach.java

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class Solution {
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public int furthestBuilding(int[] heights, int bricks, int ladders) {
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PriorityQueue<Integer> pq = new PriorityQueue<>();
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for(int i=0;i<heights.length-1;i++)
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{
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int distance = heights[i+1] - heights[i];
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if(distance>0)
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pq.add(distance);
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if(pq.size()>ladders)
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bricks -= pq.poll();
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if(bricks<0)
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return i;
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}
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return heights.length-1;
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}
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}

‎1642-furthest-building-you-can-reach/NOTES.md

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​

‎1642-furthest-building-you-can-reach/README.md

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<h2><a href="https://leetcode.com/problems/furthest-building-you-can-reach/">1642. Furthest Building You Can Reach</a></h2><h3>Medium</h3><hr><div><p>You are given an integer array <code>heights</code> representing the heights of buildings, some <code>bricks</code>, and some <code>ladders</code>.</p>
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<p>You start your journey from building <code>0</code> and move to the next building by possibly using bricks or ladders.</p>
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<p>While moving from building <code>i</code> to building <code>i+1</code> (<strong>0-indexed</strong>),</p>
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<ul>
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<li>If the current building's height is <strong>greater than or equal</strong> to the next building's height, you do <strong>not</strong> need a ladder or bricks.</li>
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<li>If the current building's height is <b>less than</b> the next building's height, you can either use <strong>one ladder</strong> or <code>(h[i+1] - h[i])</code> <strong>bricks</strong>.</li>
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</ul>
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<p><em>Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/10/27/q4.gif" style="width: 562px; height: 561px;">
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<pre><strong>Input:</strong> heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Starting at building 0, you can follow these steps:
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- Go to building 1 without using ladders nor bricks since 4 &gt;= 2.
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- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 &lt; 7.
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- Go to building 3 without using ladders nor bricks since 7 &gt;= 6.
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- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 &lt; 9.
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It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
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<strong>Output:</strong> 7
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre><strong>Input:</strong> heights = [14,3,19,3], bricks = 17, ladders = 0
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<strong>Output:</strong> 3
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= heights.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= heights[i] &lt;= 10<sup>6</sup></code></li>
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<li><code>0 &lt;= bricks &lt;= 10<sup>9</sup></code></li>
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<li><code>0 &lt;= ladders &lt;= heights.length</code></li>
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</ul>
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</div>

‎968-binary-tree-cameras/968-binary-tree-cameras.java

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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public int minCameraCover(TreeNode root) {
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int ans[] = solve(root);
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return Math.min(ans[1],ans[2]);
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}
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public int[] solve(TreeNode node)
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{
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if(node == null)
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return new int[]{0,0,9999};
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int l[] = solve(node.left);
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int r[] = solve(node.right);
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int ml = Math.min(l[1],l[2]);
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int mr = Math.min(r[1],r[2]);
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int d0 = l[1]+r[1];
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int d1 = Math.min(l[2]+mr , r[2]+ml);
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int d2 = 1+Math.min(l[0],ml) + Math.min(r[0],mr);
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return new int[]{d0,d1,d2};
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}
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}

‎968-binary-tree-cameras/NOTES.md

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​

‎968-binary-tree-cameras/README.md

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<h2><a href="https://leetcode.com/problems/binary-tree-cameras/solution/">968. Binary Tree Cameras</a></h2><h3>Hard</h3><hr><div><p>You are given the <code>root</code> of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.</p>
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<p>Return <em>the minimum number of cameras needed to monitor all nodes of the tree</em>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png" style="width: 138px; height: 163px;">
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<pre><strong>Input:</strong> root = [0,0,null,0,0]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> One camera is enough to monitor all nodes if placed as shown.
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</pre>
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<p><strong>Example 2:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png" style="width: 139px; height: 312px;">
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<pre><strong>Input:</strong> root = [0,0,null,0,null,0,null,null,0]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
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<li><code>Node.val == 0</code></li>
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</ul>
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</div>

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