MATH

Maths/CountDigit.cpp

@@ -0,0 +1,24 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int countDigits(int n)
+{
+    return (log10(n) + 1);
+}
+
+int main()
+{
+    int n;
+    cin >> n;
+    if (n == 0)
+    {
+        cout << 0;
+    }
+    else
+    {
+        n = abs(n);
+        cout << countDigits(n);
+    }
+
+    return 0;
+}

Maths/highest.cpp

@@ -0,0 +1,44 @@
+// You are given a 2D string array A of dimensions N x 2,
+
+// where each row consists of two strings: first is the name of the student, second is their marks.
+
+// You have to find the maximum average mark. If it is a floating point, round it down to the nearest integer less than or equal to the number.
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    int n;
+    cin >> n;
+    vector<vector<string>> A(n, vector<string>(2));
+    for (int i = 0; i < n; i++)
+    {
+        cin >> A[i][0] >> A[i][1];
+    }
+
+    map<string, int> m;
+    map<string, int> count;
+    for (int i = 0; i < n; i++)
+    {
+        m[A[i][0]] += stoi(A[i][1]);
+        cout << m[A[i][0]] << endl;
+        count[A[i][0]]++;
+    }
+    int max = 0;
+    string name;
+    for (auto i : m)
+    {
+        // cout << i.first << " " << i.second << endl;
+        // cout << i.first << " " << round(i.second / count[i.first]) << endl;
+        if (round (1.0 * i.second / count[i.first]) > max)
+        {
+            max = round(i.second / count[i.first]);
+            name = i.first;
+        }
+    }
+    cout << name << " " << max << endl;
+
+  return 0;
+}
+

Maths/nextSimilar.cpp

@@ -0,0 +1,38 @@
+// Given a number A in a form of string.
+
+// You have to find the smallest number that has same set of digits as A and is greater than A.
+
+// If A is the greatest possible number with its set of digits, then return -1.
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    string A;
+    cin >> A;
+    int n = A.length();
+    int i;
+    for (i = n - 1; i > 0; i--)
+    {
+        if (A[i] > A[i - 1])
+            break;
+    }
+    if (i == 0)
+    {
+        cout << -1 << endl;
+        return 0;
+    }
+    int x = A[i - 1], smallest = i;
+    for (int j = i + 1; j < n; j++)
+    {
+        if (A[j] > x && A[j] < A[smallest])
+            smallest = j;
+    }
+
+    swap(A[smallest], A[i - 1]);
+    sort(A.begin() + i, A.end());
+    cout << A << endl;
+
+    return 0;
+}

Maths/power.cpp

@@ -0,0 +1,20 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int power(int m,int n){
+    if(n==0){
+        return 1;
+
+    }
+    if(n%2==0){
+        return power(m*m,n/2);
+    }
+    return m*power(m*m,(n-1)/2);
+}
+
+int main(){
+int a,b;
+cin>>a>>b;
+cout<<pow(a, b);
+}

Maths/powerIterative.cpp

@@ -0,0 +1,30 @@
+//  2 ^ 11   ,  x^n
+//    11 ->   1 0 1 1 ;
+//  x = 2 , n=11 , res =1
+// 11/2 = 5  -> 1    res = res*x =2     n=5;    x=x*x=4
+//  5/2 = 2  -> 1    res = 2 *4 = 8     n=2     x = 16
+// 2/2 =1    -> 0    res = 8     n=1     x = 16*16 =256
+// 1/2 =0    -> 1    res = 8 *256     
+    // res = 2 * 4 *256
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int IterativePower(int x,int n){
+   int res = 1;
+    while( n ){
+        if(n & 1){
+            res = res * x ;
+        }
+        x = x * x ;
+        n = n>>1;
+    }
+    return res;
+}
+
+int main(){
+int x,n;
+cin>>x>>n;
+cout<<IterativePower(x, n);
+}

Maths/primeFactor.cpp

@@ -0,0 +1,47 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+void PrimeFactor(int n)
+{
+    if (n == 1)
+        return;
+
+    int i = 2;
+    while (n % i == 0)
+    {
+        cout << i << ' ';
+        n = n / i;
+    }
+
+    i = 3;
+    while (n % i == 0)
+    {
+        cout << i << ' ';
+        n = n / i;
+    }
+
+    for (int i = 5; i * i <= n; i = i + 6)
+    {
+        while (n % i == 0)
+        {
+            cout << i << ' ';
+            n = n / i;
+        }
+        while (n % (i + 2) == 0)
+        {
+            cout << i + 2 << ' ';
+            n = n / (i + 2);
+        }
+    }
+    if (n > 3)
+        cout << n << " ";
+}
+
+int main()
+{
+    int n;
+    cin >> n;
+    PrimeFactor(n);
+
+    return 0;
+}

Maths/primeNo.cpp

@@ -0,0 +1,30 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+bool isPrime(int n)
+{
+    if (n == 1)
+        return false;
+
+    if (n == 2 || n == 3)
+        return true;
+
+    if ((n % 2 == 0 )|| (n % 3) == 0)
+        return false;
+
+    for (int i = 5; i * i <= n; i = i + 6)
+    {
+        if ((n % i == 0) || (n % (i + 2) == 0))
+            return false;
+    }
+    return true;
+}
+
+int main()
+{
+    int n;
+    cin >> n;
+    cout << isPrime(n);
+
+    return 0;
+}

Maths/reverseInteger.cpp

@@ -0,0 +1,23 @@
+// You are given an integer N and the task is to reverse the digits of the given integer. Return 0 if the result overflows and does not fit in a 32 bit signed integer
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    int n;
+    cin >> n;
+    int rev = 0;
+    while (n != 0)
+    {
+        int rem = n % 10;
+        n /= 10;
+        if (rev > INT_MAX / 10 || (rev == INT_MAX / 10 && rem > 7))
+            return 0;
+        if (rev < INT_MIN / 10 || (rev == INT_MIN / 10 && rem < -8))
+            return 0;
+        rev = rev * 10 + rem;
+    }
+    cout << rev << endl;
+    return 0;
+}

Maths/seiveOfEratothenes.cpp

@@ -0,0 +1,22 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+void printAllPrime(int n){
+    vector<bool> isPrime(n+1,true);
+    for(int i=2; i<=n ; i++){
+        if(isPrime[i]){
+            cout<<i<<" ";
+            for(int j= i*i; j<=n; j=j+i){
+                isPrime[j] = false;
+            }
+        }
+    }
+
+}
+
+int main(){
+    int n;
+    cin>>n;
+    printAllPrime(n);
+    return 0;
+}

Maths/trailingZeroInFactorial.cpp

@@ -0,0 +1,21 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int countTrailingZeroes(int n)
+{
+   int c =1;
+   int countZeroes = 0;
+   while((n/(pow(5,c)))){
+       countZeroes += (n/(pow(5,c)));
+       c++;
+   }
+   return countZeroes;
+}
+
+int main()
+{
+    int n;
+    cin >> n;
+    cout << countTrailingZeroes(n);
+    return 0;
+}