Initial Files

Author: arya-io

PRACTICE/10_carry.cpp

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+/*
+Question:10
+(Asked in Accenture Offcampus 1 Aug 2021, Slot 2)
+
+Problem Statement
+
+A carry is a digit that is transferred to left if sum of digits exceeds 9 while adding two numbers from right-to-left one digit at a time
+
+You are required to implement the following function.
+
+Int NumberOfCarries(int num1 , int num2);
+
+The functions accepts two numbers ‘num1’ and ‘num2’ as its arguments. You are required to calculate and return  the total number of carries generated while adding digits of two numbers ‘num1’ and ‘ num2’.
+
+Assumption: num1, num2>=0
+
+Example:
+
+Input
+Num 1: 451
+Num 2: 349
+Output
+2
+Explanation:
+
+Adding ‘num 1’ and ‘num 2’ right-to-left results in 2 carries since ( 1+9) is 10. 1 is carried and (5+4=1) is 10, again 1 is carried. Hence 2 is returned.
+
+Sample Input
+
+Num 1: 23
+
+Num 2: 563
+
+Sample Output
+
+0
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int numberOfCarries(int num1, int num2){
+
+    int greater, count = 0, rem1 = 0, rem2 = 0, carry = 0;
+
+    (num1 > num2 ? greater = num1 : greater = num2);
+
+    while(greater){
+        rem1 = num1 % 10;
+        rem2 = num2 % 10;
+
+        if(rem1 + rem2 + carry > 9) count++;
+
+        carry = (rem1+rem2)/10;
+
+        greater /= 10;
+        num1 /= 10;
+        num2 /= 10;
+    }
+
+    return count;
+}
+
+int main(){
+    int num1, num2;
+    cin >> num1 >> num2;
+    cout << numberOfCarries(num1, num2);
+}

PRACTICE/11_replace.cpp

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+/*
+Question:11
+(Asked in Accenture Offcampus 1 Aug 2021, Slot 3)
+
+Problem Statement
+
+You are given a function,
+
+Void *ReplaceCharacter(Char str[], int n, char ch1, char ch2);
+
+The function accepts a string  ‘ str’ of length n and two characters ‘ch1’ and ‘ch2’ as its arguments . Implement the function to modify and return the string ‘ str’ in such a way that all occurrences of ‘ch1’ in original string are replaced by ‘ch2’ and all occurrences of ‘ch2’  in original string are replaced by ‘ch1’.
+
+Assumption: String Contains only lower-case alphabetical letters.
+
+Note:
+
+Return null if string is null.
+If both characters are not present in string or both of them are same , then return the string unchanged.
+Example:
+
+Input:
+Str: apples
+ch1:a
+ch2:p
+Output:
+paales
+Explanation:
+
+‘A’ in original string is replaced with ‘p’ and ‘p’ in original string is replaced with ‘a’, thus output is paales.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+string replaceCharacter(string s, int n, char c1, char c2){
+    if(n == 0) return 0;
+
+    for(int i = 0; i < n; i++){
+        if(s[i] == c1) s[i] = c2;
+        else if(s[i] == c2) s[i] = c1;
+    }
+
+    return s;
+}
+
+int main(){
+    string s;
+    char c1, c2;
+    cin >> s >> c1 >> c2;
+    int n = s.size();
+    cout << replaceCharacter(s, n, c1, c2);
+}

PRACTICE/12operations.cpp

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+/*
+Question:12
+(Asked in Accenture Offcampus 2 Aug 2021, Slot 1)
+
+Problem Statement
+
+You are required to implement the following function.
+
+Int OperationChoices(int c, int n, int a , int b )
+
+The function accepts 3 positive integers ‘a’ , ‘b’ and ‘c ‘ as its arguments. Implement the function to return.
+
+( a+ b ) , if c=1
+( a – b ) , if c=2
+( a * b ) ,  if c=3
+(a / b) ,  if c =4
+Assumption : All operations will result in integer output.
+
+Example:
+
+Input
+c :1
+a:12
+b:16
+Output:
+Since ‘c’=1 , (12+16) is performed which is equal to 28 , hence 28 is returned.
+Sample Input
+
+ c : 2
+
+ a : 16
+
+ b : 20
+
+Sample Output
+
+-4
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int operationChoices(int c, int a, int b){
+    if(c == 1) return a+b;
+    if(c == 2) return a-b;
+    if(c == 3) return a*b;
+    if(c == 4) return a/b;
+
+    return 0;
+}
+
+int main(){
+    int a, b, c;
+    cin >> c >> a >> b;
+    cout << operationChoices(c, a, b);
+}

PRACTICE/13_exponent.cpp

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+/*
+Question:13
+(Asked in Accenture Offcampus 2 Aug 2021, Slot 2)
+
+Problem Statement
+
+You are given a function,
+
+Int MaxExponents (int a , int b);
+
+You have to find and return the number between ‘a’ and ‘b’ ( range inclusive on both ends) which has the maximum exponent of 2.
+
+The algorithm to find the number with maximum exponent of 2 between the given range is
+
+Loop between ‘a’ and ‘b’. Let the looping variable be ‘i’.
+Find the exponent (power) of 2 for each ‘i’ and store the number with maximum exponent of 2 so faqrd in a variable , let say ‘max’. Set ‘max’ to ‘i’ only if ‘i’ has more exponent of 2 than ‘max’.
+Return ‘max’.
+Assumption: a <b
+
+Note: If two or more numbers in the range have the same exponents of  2 , return the small number.
+
+Example
+
+Input:
+7
+12
+Output:
+8
+Explanation:
+
+Exponents of 2 in:
+
+7-0
+
+8-3
+
+9-0
+
+10-1
+
+11-0
+
+12-2
+
+Hence maximum exponent if two is of 8.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int maxExponents(int a, int b){
+    int ans = 0, max = 0, currenti;
+
+    for(int i = a; i <= b; i++){
+        int count = 1;
+        int ci = i;
+        if(ci%2==0){
+            while(ci%2==0){
+                count++;
+                ci = ci/2;
+            }
+            if(count>max){
+                max = count;
+                ans = i;
+
+            }
+
+        }
+
+    }
+    return ans;
+
+    return ans;
+}
+
+int main(){
+    int a, b;
+    cin >> a >> b;
+    cout << maxExponents(a, b);
+}

PRACTICE/14_calculate.cpp

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+/*
+
+Question : 14
+(Asked in Accenture Offcampus 2 Aug 2021, Slot 3)
+
+Problem Statement
+
+You are required to implement the following function:
+
+Int Calculate(int m, int n);
+
+The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.You are required to calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive and return the same.
+Note
+0 < m <= n
+
+Example
+
+Input:
+
+m : 12
+
+n : 50
+
+Output
+
+90
+
+Explanation:
+The numbers divisible by both 3 and 5, between 12 and 50 both inclusive are {15, 30, 45} and their sum is 90.
+Sample Input
+m : 100
+n : 160
+Sample Output
+510
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int calculate (int m, int n){
+
+    int sum = 0;
+
+    for(int i = m; i <= n;){
+        if(i%3 == 0 && i % 5 == 0){
+            sum += i;
+            i += 15;
+        }
+        else i++;
+    }
+
+    return sum;
+}
+
+int main(){
+    int m, n;
+    cin >> m >> n;
+
+    cout << calculate(m, n);
+}

PRACTICE/15_largest.cpp

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+/*
+
+Question 15
+Problem Statement 
+
+You are required to input the size of the matrix then the elements of matrix, then you have to divide the main matrix in two sub matrices (even and odd) in such a way that element at 0 index will be considered as even and element at 1st index will be considered as odd and so on. then you have sort the even and odd matrices in ascending order then print the sum of second largest number from both the matrices
+
+Example
+
+enter the size of array : 5
+enter element at 0 index : 3
+enter element at 1 index : 4
+enter element at 2 index : 1
+enter element at 3 index : 7
+enter element at 4 index : 9
+Sorted even array : 1 3 9
+Sorted odd array : 4 7
+
+7
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int largest(int n, int arr[]){
+    vector<int> even;
+    vector<int> odd;
+
+    for(int i = 0; i < n; i++){
+        if(i&1) odd.push_back(arr[i]);
+        else even.push_back(arr[i]);
+    }
+
+    sort(odd.begin(), odd.end(), greater<int>());
+    sort(even.begin(), even.end(), greater<int>());
+
+    return(odd[1]+even[1]);
+}
+
+int main(){
+    int n;
+    cin >> n;
+    int arr[n];
+    for(int i =0; i < n; i++) cin >> arr[i];
+
+    cout << largest(n, arr) << endl;
+}