ARRAYS

Arrays/Break Words.cpp

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+/*
+You are given a sentence contained in given string S. Write a function which will replace all the words within the sentence whose length is even and greater than equal to 4, with a space between the two equal halves of the word.
+Space complexity should be O(1).
+Input Format :
+
+String S
+
+Output Format :
+
+Updated string
+
+Constraints :
+1 <= Length of S <= 10000
+Sample Input :
+
+Helloo world good morniing
+
+Sample Output :
+
+Hel loo world go od morn iing
+
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+void splitWord(string &word){
+    int mid=word.size()/2;
+    word.insert(mid," ");
+}
+
+void breakWords(char* S)
+{
+	/*
+	 * Don't write main.
+	 * Don't return or print anything.
+	 * Changes should be done in the given string.
+	*/
+    string ans="";
+    string word="";
+    for(int i=0;S[i]!='\0';i++){
+
+        if(S[i]!=' ' && S[i]!='\0')
+            word+=S[i];
+        else{
+            if (word.size()>=4 && word.size()%2==0){
+                splitWord(word);
+            }
+            ans+=word+" ";
+            word="";
+        }
+    }
+    strcpy(S,ans.c_str());
+
+}
+
+
+#include<iostream>
+#include<string.h>
+using namespace std;
+#include"solution.h"
+int main()
+{
+	char str[100000];
+	cin.getline(str,100000);
+	breakWords(str);
+	cout<<str;
+}

Arrays/Check_Permuations.cpp

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+/*
+Given two strings, check if they are permutations of each other. Return true or false.
+Permutation means - length of both the strings should same and should contain same set of characters. Order of characters doesn't matter.
+Note : Input strings contain only lowercase english alphabets.
+Input format :
+
+Line 1 : String 1
+Line 2 : String 2
+
+Sample Input 1 :
+
+abcde
+baedc
+
+Sample Output 1 :
+
+true
+
+Sample Input 2 :
+
+abc
+cbd
+
+Sample Output 2 :
+
+false
+
+
+*/
+
+// input1 - first input string
+// input2 - second input string
+#include <bits/stdc++.h>
+using namespace std;
+
+bool isPermutation(char input1[], char input2[]) {
+    // Write your code here
+    int size1=strlen(input1),size2=strlen(input2);
+    sort(input1,input1+size1);
+    sort(input2,input2+size2);
+
+    if(!strcmp(input1,input2))
+        return true;
+    else
+        return false;
+
+}
+
+#include <iostream>
+using namespace std;
+#include "solution.h"
+
+int main() {
+    char input1[1000], input2[1000];
+    cin.getline(input1, 1000);
+    cin.getline(input2, 1000);
+    if(isPermutation(input1, input2) == 1) {
+        cout << "true";
+    }
+    else {
+        cout << "false";
+    }
+}
+

Arrays/Compress_the_String.cpp

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+/*
+Write a program to do basic string compression. For a character which is consecutively repeated more than once, replace consecutive duplicate occurrences with the count of repetitions.
+For e.g. if a String has 'x' repeated 5 times, replace this "xxxxx" with "x5".
+Note : Consecutive count of every character in input string is less than equal to 9.
+Input Format :
+
+Input string S
+
+Output Format :
+
+Compressed string 
+
+Sample Input:
+
+aaabbccdsa
+
+Sample Output:
+
+a3b2c2dsa
+
+
+*/
+
+// input - given string
+// You need to update in the given string itself i.e. in input. No need to return or print.
+#include <bits/stdc++.h>
+using namespace std;
+
+void stringCompression(char input[]) {
+    // Write your code here
+    string ans="";
+
+    for(int i=0;input[i]!=0;){
+        int j=i+1;
+        while(input[i]==input[j]){
+            j++;
+        }
+        int c=j-i;
+        //cout<<input[i]<<endl;
+        //cout<<char(48+c)<<endl;
+        //cout<<ans<<endl;
+        char cd='0'+c;
+        if (c!=0 && c!=1){
+            ans+=input[i];
+            ans+=(char)(cd);
+        }
+        else
+            ans+=input[i];
+        i=j;
+    }
+    strcpy(input,ans.c_str());
+
+}
+
+#include <iostream>
+#include "solution.h"
+using namespace std;
+
+int main() {
+    char input[1000];
+    cin.getline(input, 1000);
+    stringCompression(input);
+    cout << input << endl;
+}

Arrays/Count Platforms.cpp

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+/*
+Given two arrays (both of size n), one containing arrival time of trains and other containing the corresponding trains departure time (in the form of an integer) respectively. Return the minimum number of platform required, such that no train waits.
+Arrival and departure time of a train can't be same.
+Input Format :
+
+Line 1: Integer n, number of elements in arrival and departure array
+Line 2: Elements of Arrival Array (separated by spaces).
+Line 3: Elements of Departure Array (separated by spaces).
+
+Output Format:
+
+Minimum Number of Platform Needed
+
+Constraints :
+1 <= n <= 100
+Sample Input 1 :
+
+6
+900 940 950 1100 1500 1800
+910 1200 1120 1130 1900 2000
+
+Sample Output 1 :
+
+3
+
+Sample Input 2 :
+
+4
+1100 1101 1103 1105
+1110 1102 1104 1106
+
+Sample Output 2 :
+
+2
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+int platformsNeeded(int arrival[], int departure[], int n) {
+    /*
+     * Don't write main().
+     * Don't read input, it is passed as function argument.
+     * Don't print anything just return integer value.
+     */
+    int maxc=1;
+    int c=1;
+    int i=0,j=0;
+
+    sort(arrival,arrival+n);
+    sort(departure,departure+n);
+
+    while(i<n && j<n){
+        if(arrival[i]<departure[j]){
+            i++;
+            c++;
+        }
+        else{
+            j++;
+            c--;
+            maxc=max(maxc,c);
+        }
+    }
+    return maxc;
+}
+
+#include<iostream>
+using namespace std;
+#include"solution.h"
+
+int main()
+{
+	int n;
+	cin>>n;
+    int* arr=new int[n];
+    int* dep=new int[n];
+    for(int i=0;i<n;i++)
+    {
+    	cin>>arr[i];
+	}
+	for(int i=0;i<n;i++)
+    {
+    	cin>>dep[i];
+	}
+    cout<< platformsNeeded(arr, dep, n);
+}

Arrays/Find_Duplicate.cpp

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+#include <iostream>
+using namespace std;
+#include "solution.h"
+
+
+// arr - input array
+// size - size of array
+
+int MissingNumber(int arr[], int size){
+    /* Don't write main().
+     * Don't read input, it is passed as function argument.
+     * Return output and don't print it.
+     * Taking input and printing output is handled automatically.
+     */
+    int sum=(size*(size-1))/2;
+
+    int calsum=0;
+    for(int i=0;i<size;i++)
+        calsum+=arr[i];
+    int k=sum-calsum;
+    return size-1-k;
+
+
+}
+
+
+int main() {
+	int size;
+	cin >> size;
+	int *input = new int[1 + size];
+
+	for(int i = 0; i < size; i++)
+		cin >> input[i];
+
+	cout << MissingNumber(input, size);	
+
+	delete [] input;
+
+	return 0;
+}

Arrays/Find_an_element.cpp

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+#include <iostream>
+using namespace std;
+
+
+void findElement(int input[][100], int m, int n, int x) {
+	int i = 0, j = n - 1;
+	while(i < m && j >= 0) {
+		if(input[i][j] == x) {
+			cout << i << " " << j << endl;
+			return;
+		}
+		else if(x > input[i][j]) {
+			i++;
+		}
+		else {
+			j--;
+		}
+	}
+	cout << "-1" << endl;
+}
+
+void printArray(int input[][100], int m, int n) {
+	for(int i = 0; i < m; i++) {
+		for(int j = 0;  j < n; j++) {
+			cout << input[i][j] << " ";
+		}
+		cout << endl;	
+	}
+
+}
+
+int main() {
+	int a[100][100];
+	int m, n;
+
+	// Rows - m
+	// Cols - n
+	cin >> m >> n;
+
+
+	// Taking input
+	for(int i = 0; i < m; i++) {
+		for(int j = 0; j < n; j++) {
+			cin >> a[i][j];
+		}
+	}
+
+	int x;
+	cin >> x;
+
+	findElement(a, m, n, x);
+
+	//printArray(a, m, n);
+
+}
+