🎯Added cpp solution Q5 in searching and sorting.

Love Babbar DSA Sheet Solutions/Searching & Sorting/Find Missing and Repeating/find_missing_and_repeating.cpp

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+//Problem Statement : 
+/* 
+Given an unsorted array Arr of size N of positive integers. One number 'A' from set {1, 2, …N} is missing and one number 'B' occurs twice in array. Find these two numbers.
+
+Example 1:
+
+Input:
+N = 2
+Arr[] = {2, 2}
+Output: 2 1
+Explanation: Repeating number is 2 and 
+smallest positive missing number is 1.
+Example 2:
+
+Input:
+N = 3
+Arr[] = {1, 3, 3}
+Output: 3 2
+Explanation: Repeating number is 3 and 
+smallest positive missing number is 2.
+*/
+
+//Function to solve the problem
+int *findTwoElement(int *arr, int n) {
+ int *ans = new int[2];
+ unordered_map<int, int> mp;
+ for(int i = 0; i < n; i++)
+ mp[arr[i]]++ ;
+
+
+ for(int i = 1; i < n + 1; i++)
+ if(mp[arr[i]] > 1)
+ ans[0] = arr[i];
+ // m n 
+ for(int i = 1; i <= n; i++)
+ if(mp[i] == 0){
+ ans[1] = i;
+ break;
+ }
+
+ return ans; 
+ }

Love Babbar DSA Sheet Solutions/Searching & Sorting/First And Last Occurrences of x/first_and_last_occurrences_of_x.cpp

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+//Problem Statement : 
+/* 
+Given a sorted array arr containing n elements with possibly duplicate elements, the task is to find indexes of first 
+and last occurrences of an element x in the given array.
+
+Difficulty : Basic
+*/
+
+
+/*
+Example 1:
+Input:
+n=9, x=5
+arr[] = { 1, 3, 5, 5, 5, 5, 67, 123, 125 }
+Output: 2 5
+Explanation: First occurrence of 5 is at index 2 and last occurrence of 5 is at index 5. 
+
+
+Example 2:
+Input:
+n=9, x=7
+arr[] = { 1, 3, 5, 5, 5, 5, 7, 123, 125 }
+Output: 6 6 
+
+
+/*
+Expected Time Complexity: O(logN)
+Expected Auxiliary Space: O(1).
+*/
+
+
+// Solution Explained -
+/* 
+ 1) Iterate the array from both forward and backward directions with 2 pointers.
+ 2) As soon as we find the required element in the array we will push_back the respective indices(values of those 2 pointers) in a vector.
+ 3) Return the vector.
+
+*/
+
+
+ //Solution--
+
+// { Driver Code Starts
+#include<bits/stdc++.h>
+using namespace std;
+vector<int> find(int a[], int n , int x );
+
+int main()
+{
+ int t;
+ cin>>t;
+ while(t--)
+ {
+ int n,x;
+ cin>>n>>x;
+ int arr[n],i;
+ for(i=0;i<n;i++)
+ cin>>arr[i];
+ vector<int> ans;
+ ans=find(arr,n,x);
+ cout<<ans[0]<<" "<<ans[1]<<endl;
+ }
+ return 0;
+}
+// } Driver Code Ends
+
+vector<int> find(int arr[], int n , int x )
+{
+ int i = 0, j = n -1;
+ vector<int> v;
+ while(i <= j){
+ if(arr[i] != x)
+ i++ ;
+
+ if(arr[j] != x)
+ j-- ;
+
+ if(arr[i] == x && arr[j] == x){
+ v.push_back(i);
+ v.push_back(j);
+ break;
+ } 
+ }
+ if(i > j){
+ v.push_back(-1);
+ v.push_back(-1);
+ }
+
+ return v;
+}

Love Babbar DSA Sheet Solutions/Searching & Sorting/Search in Rotated Sorted Array/search_in_rotated_sorted_array.cpp

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+//LEETCODE PROBLEM
+
+//Problem Statement : 
+/* 
+There is an integer array nums sorted in ascending order (with distinct values).
+
+Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting 
+array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at 
+pivot index 3 and become [4,5,6,7,0,1,2].
+
+Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+
+You must write an algorithm with O(log n) runtime complexity.
+
+Difficulty : Medium
+*/
+
+
+/*
+Example 1:
+
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1
+Example 3:
+
+Input: nums = [1], target = 0
+Output: -1
+*/
+
+//Solution-
+class Solution {
+public:
+ int search(vector<int>& nums, int target) {
+ int s=0;
+ int e=nums.size()-1;
+ while(s<=e)
+ {
+ int m = s + (e-s)/2;
+ if(nums[m]==target || nums[s]==target)
+ { 
+ if(nums[s]==target) return s;
+
+ return m;
+ }
+ if((target<nums[s]&&target<nums[m])||(target>nums[s]&&target>nums[m]))
+ {
+ if(nums[s]>nums[m])
+ e = m-1;
+
+ else
+ s = m+1;
+ }
+ else
+ {
+ if(nums[s]>nums[m])
+ s = m+1;
+
+ else
+ e = m-1;
+ }
+ }
+ return -1;
+ } 
+};

Love Babbar DSA Sheet Solutions/Searching & Sorting/Value equal to index value/value_equal_to_index.cpp

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+//Problem Statement : 
+/* 
+Given an array Arr of N positive integers. Your task is to find the elements whose value 
+is equal to that of its index value ( Consider 1-based indexing ).
+
+Difficulty : School
+*/
+
+
+/*
+Example 1:
+Input: 
+N = 5
+Arr[] = {15, 2, 45, 12, 7}
+Output: 2
+Explanation: Only Arr[2] = 2 exists here.
+*/
+
+
+//Solution Explained
+//Easy School level problem. Added just for the sake of sequence...
+
+
+// Solution-
+
+#include<bits/stdc++.h>
+using namespace std;
+
+class Solution{
+public:
+vector<int> valueEqualToIndex(int arr[], int n) {
+ int i = 0;
+ vector<int> v;
+ while(i <= n){
+ if(arr[i] == i+1){
+ v.push_back(arr[i]);
+ i++ ;
+ }
+ else
+ i++ ; 
+ }
+ return v;
+}
+};
+
+// { Driver Code Starts.
+int main() {
+ int t;
+ cin >> t;
+ while (t--) {
+ int n, i;
+ cin >> n;
+ int arr[n];
+ for (i = 0; i < n; i++) {
+ cin >> arr[i];
+ }
+ Solution ob;
+ auto ans = ob.valueEqualToIndex(arr, n);
+ if (ans.empty()) {
+ cout << "Not Found";
+ } else {
+ for (int x: ans) {
+ cout << x << " ";
+ }
+ }
+ cout << "\n";
+ }
+ return 0;
+}
+ // } Driver Code Ends