added new problems

Added c++ Solutions of leetcode

Co-Authored-By: Vaibhav Nirmal <[email protected]>

01-matrix/README.md

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+<h2>542. 01 Matrix</h2><h3>Medium</h3><hr><div><p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>
+
+<p>The distance between two adjacent cells is <code>1</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;">
+<pre><strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
+<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;">
+<pre><strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
+<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == mat.length</code></li>
+	<li><code>n == mat[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
+	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
+	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
+</ul>
+</div>

1's Complement - GFG/1s-complement.cpp

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+// { Driver Code Starts
+
+#include<bits/stdc++.h>
+using namespace std; 
+
+ // } Driver Code Ends
+
+class Solution{ 
+public:
+     string onesComplement(string S,int N){
+       //code here
+   int i=0;
+   string result="";
+   while(i<N){
+       if(S[i]=='1'){
+           result.append("0");
+       }
+       else{
+           result.append("1");
+           }
+           i++;
+       }
+       return result;
+
+   }
+};
+
+// { Driver Code Starts.
+int main()
+{
+    int t;
+    cin>>t;
+    while(t--)
+    {
+        int n;
+        cin>>n;
+        string s;
+        cin>>s;
+        Solution ob;
+        cout<<ob.onesComplement(s,n)<<"\n";
+    }
+}  // } Driver Code Ends

1's Complement - GFG/README.md

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+# 1's Complement
+## Easy
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an<strong> N </strong>bit binary number, find the 1's complement of the number.&nbsp;The ones'&nbsp;complement&nbsp;of a binary number is&nbsp;defined&nbsp;as the value obtained by inverting all the bits in the binary representation of the number (swapping 0s for 1s and vice versa).</span><br>
+&nbsp;</p>
+
+<p><strong><span style="font-size:18px">Example 1:</span></strong></p>
+
+<pre><strong><span style="font-size:18px">Input:</span>
+</strong><span style="font-size:18px">N = 3
+S = 101
+<strong>Output:
+</strong>010
+<strong>Explanation:
+</strong>We get the output by converting 1's in S
+to 0 and 0s to 1</span><strong>
+</strong></pre>
+
+<p><strong><span style="font-size:18px">Example 2:</span></strong></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong>
+N = 2
+S = 10
+<strong>Output:</strong>
+01
+<strong>Explanation:</strong>
+We get the output by converting 1's in S
+to 0 and 0s to 1</span>
+</pre>
+
+<p><br>
+<span style="font-size:18px"><strong>Your Task:&nbsp;&nbsp;</strong><br>
+You don't need to read input or print anything. Your task is to complete the function&nbsp;<strong>onesComplement()</strong>&nbsp;which takes the binary string S, its size N<strong>&nbsp;</strong>as input parameters&nbsp;and returns 1's complement of S of size N.</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity:</strong> O(N)<br>
+<strong>Expected Space Complexity:</strong> O(N)</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1&lt;=N&lt;=100</span></p>
+ <p></p>
+            </div>

1-two-sum/1-two-sum.cpp

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+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        unordered_map <int,int> m;
+        vector <int> v;
+
+        for(int i=0;i<nums.size();i++)
+        {
+            if(m.find(target - nums[i]) != m.end())
+            {
+                v.push_back(i);
+                v.push_back(m[target-nums[i]]);
+            }
+            else
+            {
+                m[nums[i]] = i;
+            }
+        }
+
+        return v;
+    }
+};

1-two-sum/NOTES.md

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+​

1-two-sum/README.md

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+<h2><a href="https://leetcode.com/problems/two-sum/">1. Two Sum</a></h2><h3>Easy</h3><hr><div><p>Given an array of integers <code>nums</code>&nbsp;and an integer <code>target</code>, return <em>indices of the two numbers such that they add up to <code>target</code></em>.</p>
+
+<p>You may assume that each input would have <strong><em>exactly</em> one solution</strong>, and you may not use the <em>same</em> element twice.</p>
+
+<p>You can return the answer in any order.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [2,7,11,15], target = 9
+<strong>Output:</strong> [0,1]
+<strong>Explanation:</strong> Because nums[0] + nums[1] == 9, we return [0, 1].
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,2,4], target = 6
+<strong>Output:</strong> [1,2]
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,3], target = 6
+<strong>Output:</strong> [0,1]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
+	<li><strong>Only one valid answer exists.</strong></li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow-up:&nbsp;</strong>Can you come up with an algorithm that is less than&nbsp;<code>O(n<sup>2</sup>)&nbsp;</code>time complexity?</div>

100-same-tree/NOTES.md

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+​

100-same-tree/README.md

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+<h2><a href="https://leetcode.com/problems/same-tree/">100. Same Tree</a></h2><h3>Easy</h3><hr><div><p>Given the roots of two binary trees <code>p</code> and <code>q</code>, write a function to check if they are the same or not.</p>
+
+<p>Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/12/20/ex1.jpg" style="width: 622px; height: 182px;">
+<pre><strong>Input:</strong> p = [1,2,3], q = [1,2,3]
+<strong>Output:</strong> true
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/12/20/ex2.jpg" style="width: 382px; height: 182px;">
+<pre><strong>Input:</strong> p = [1,2], q = [1,null,2]
+<strong>Output:</strong> false
+</pre>
+
+<p><strong>Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/12/20/ex3.jpg" style="width: 622px; height: 182px;">
+<pre><strong>Input:</strong> p = [1,2,1], q = [1,1,2]
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in both trees is in the range <code>[0, 100]</code>.</li>
+	<li><code>-10<sup>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
+</ul>
+</div>

1009-complement-of-base-10-integer/1009-complement-of-base-10-integer.cpp

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+class Solution {
+public:
+  int bitwiseComplement(int N, int c = 1) {
+  while (c < N) c = (c << 1) + 1;
+  return N ^ c;
+}
+};

1009-complement-of-base-10-integer/NOTES.md

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+​

1009-complement-of-base-10-integer/README.md

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+<h2><a href="https://leetcode.com/problems/complement-of-base-10-integer/">1009. Complement of Base 10 Integer</a></h2><h3>Easy</h3><hr><div><p>The <strong>complement</strong> of an integer is the integer you get when you flip all the <code>0</code>'s to <code>1</code>'s and all the <code>1</code>'s to <code>0</code>'s in its binary representation.</p>
+
+<ul>
+	<li>For example, The integer <code>5</code> is <code>"101"</code> in binary and its <strong>complement</strong> is <code>"010"</code> which is the integer <code>2</code>.</li>
+</ul>
+
+<p>Given an integer <code>n</code>, return <em>its complement</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> n = 5
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> n = 7
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre><strong>Input:</strong> n = 10
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= n &lt; 10<sup>9</sup></code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Note:</strong> This question is the same as 476: <a href="https://leetcode.com/problems/number-complement/" target="_blank">https://leetcode.com/problems/number-complement/</a></p>
+</div>

1010-pairs-of-songs-with-total-durations-divisible-by-60/1010-pairs-of-songs-with-total-durations-divisible-by-60.cpp

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+class Solution {
+public:
+    int numPairsDivisibleBy60(vector<int>& time) {
+        vector<long long int> modd(60);
+        long long int total = 0;
+        for(auto a:time) modd[a%60]++;
+        for(int i=1;i<=29;i++) total += (modd[i] * modd[60-i]);
+        total = total + modd[30]*(modd[30]-1)/2 ;
+        total += modd[0]*(modd[0]-1)/2;
+        return total;
+    }
+};

1010-pairs-of-songs-with-total-durations-divisible-by-60/NOTES.md

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+​

1010-pairs-of-songs-with-total-durations-divisible-by-60/README.md

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+<h2><a href="https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/">1010. Pairs of Songs With Total Durations Divisible by 60</a></h2><h3>Medium</h3><hr><div><p>You are given a list of songs where the <code>i<sup>th</sup></code> song has a duration of <code>time[i]</code> seconds.</p>
+
+<p>Return <em>the number of pairs of songs for which their total duration in seconds is divisible by</em> <code>60</code>. Formally, we want the number of indices <code>i</code>, <code>j</code> such that <code>i &lt; j</code> with <code>(time[i] + time[j]) % 60 == 0</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> time = [30,20,150,100,40]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> Three pairs have a total duration divisible by 60:
+(time[0] = 30, time[2] = 150): total duration 180
+(time[1] = 20, time[3] = 100): total duration 120
+(time[1] = 20, time[4] = 40): total duration 60
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> time = [60,60,60]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> All three pairs have a total duration of 120, which is divisible by 60.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= time.length &lt;= 6 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= time[i] &lt;= 500</code></li>
+</ul>
+</div>

1011-capacity-to-ship-packages-within-d-days/1011-capacity-to-ship-packages-within-d-days.cpp

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+class Solution {
+public:
+    bool isPossible(int mid, vector<int>& nums, int days)
+    {
+        int count = 1;
+        int curr_sum = 0;
+        for(int i=0;i<nums.size();i++)
+        {
+            if(curr_sum + nums[i] > mid)
+            {
+                count++;
+                if(count > days)
+                    return false;
+
+                curr_sum = nums[i];
+            }
+            else
+            {
+                curr_sum += nums[i];
+            }
+        }
+        cout<<"count  mid "<<count<<" "<<mid<<endl;
+        return true;
+    }
+    int shipWithinDays(vector<int>& nums, int days) {
+
+        int low = 0;
+        int high = 0;
+
+        for(int i=0;i<nums.size();i++)
+        {
+            low = max(low,nums[i]);
+            high += nums[i];
+        }
+        int ans = 0;
+        while(low <= high)
+        {
+            int mid = (low+high)>>1;
+            //cout<<mid<<" ";
+            if(isPossible(mid,nums,days))
+            {
+
+                ans = mid;
+                high = mid - 1;
+            }
+            else
+                low = mid + 1; 
+        }
+        return ans;
+    }
+};

1011-capacity-to-ship-packages-within-d-days/NOTES.md

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+​