Merge pull request #591 from geeky-auro/main

Some Important Array Problems Added

Author: gantavyamalviya

Arrays/32_Leaders_in_Array.cpp

@@ -0,0 +1,57 @@
+/*
+Given an integer array A of size n. 
+Find and print all the leaders present in the input array. 
+An array element A[i] is called Leader, if all the elements following it (i.e. present at its right) are less than or equal to A[i]. 
+Print all the leader elements separated by space and in the same order they are present in the input array. 
+
+Input Format : 
+Line 1 : Integer n, size of array 
+Line 2 : Array A elements (separated by space) 
+Output Format : leaders of array (separated by space) 
+
+Constraints : 1 <= n <= 10^6 
+Sample Input 1 : 6 3 12 34 2 0 -1 
+Sample Output 1 : 34 2 0 -1 
+
+Sample Input 2 : 5 13 17 5 4 6 
+Sample Output 2 : 17 6
+*/
+#include<iostream>
+#include<climits>
+using namespace std;
+void Leaders(int* arr,int len)
+{
+   for(int i=0;i<len;i++){
+        bool isbig=true; //Initially assume that present element is greater than next element
+    for(int j=i+1;j<len;j++){
+        // check for the next element if it is greater change then change flag "isbig" to false
+        if(!(arr[i]>=arr[j])){
+            isbig=false;
+            break;
+        }
+        // if they are arranged as they should be as in the case of leaders then change the flag to true 
+        // Or you can leave the flag unchanged as it is ....!
+        // In the below codes are added for better understanding and Workflow.
+        else{
+            isbig=true;
+            continue;
+        }
+    }
+    if(isbig){
+        // Print the element for Leaders..!
+        cout<<arr[i]<<" ";
+    }
+}
+}
+int main()
+{
+    int len;
+    cin>>len;
+	int *arr = new int[len + 1];
+	
+	for(int i=0;i<len;i++)
+	{
+		cin>>arr[i];
+	}
+	Leaders(arr,len);
+}

Arrays/33_Sort_0_1_2.cpp

@@ -0,0 +1,81 @@
+/*
+
+You are given an integer array/list(ARR) of size N. 
+It contains only 0s, 1s and 2s. 
+Write a solution to sort this array/list in a 'single scan'. 
+'Single Scan' refers to iterating over the array/list just once or to put it in other words, you will be visiting each element in the array/list just once. 
+Note: You need to change in the given array/list itself. 
+Hence, no need to return or print anything.
+
+Input format : The first line contains an Integer 't' which denotes the number of test cases or queries to be run. 
+Then the test cases follow.
+
+First line of each test case or query contains an integer 'N' representing the size of the array/list.
+
+Second line contains 'N' single space separated integers(all 0s, 1s and 2s) representing the elements in the array/list. Output Format : For each test case, print the sorted array/list elements in a row separated by a single space.
+
+Output for every test case will be printed in a separate line. 
+Constraints : 
+1 <= t <= 10^2 0 <= N <= 10^5 
+Time Limit: 1 sec
+
+Sample Input 1: 1 7 0 1 2 0 2 0 1 
+Sample Output 1: 0 0 0 1 1 2 2
+
+Sample Input 2: 2 5 2 2 0 1 1 7 0 1 2 0 1 2 0 
+Sample Output 2: 0 1 1 2 2 0 0 0 1 1 2 2
+
+*/
+#include<iostream>
+using namespace std;
+
+void simplifiedsort012(int *arr, int n){
+
+    int i=0,nz=0,nt=n-1;
+    while(i<n &&i<=nt){
+        if(arr[i]==0){
+              int temp=arr[i];
+            arr[i]=arr[nz];
+            arr[nz]=temp;
+            nz++;
+        }
+        else if(arr[i]==2){
+                int temp=arr[i];
+            arr[i]=arr[nt];
+            arr[nt]=temp;
+            nt--;
+            continue;
+
+        }
+
+        i++;
+    }
+}
+
+int main(){
+
+    int n;
+    cout<<"Enter the size of the Array "<<endl;
+    cin>>n;
+    int *arr=new int[n];
+    cout<<"Enter Elements in the Array "<<endl;
+    for(int i=0;i<n;i++){
+        cin>>arr[i];
+    }
+    simplifiedsort012(arr,n);
+    return 0;
+}
+
+/*
+Explanation
+👉🏼Take two variables as starting(nz) and ending(nt) depicting position of 0's and 2's to be swapped respectively
+
+👉🏼Initialize nz as Starting Index i.e 0 and increment it after every swap
+
+👉🏼Initialize nt as Ending Index i.e Size-1 and decrement it after every swap
+
+Special Attention should be given to the following points :)
+✨Swap 0's and increment nz & i
+
+✨Swap 2's and increment nt but not i (ignore the updation of expression)
+*/