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Author: adityajamwal02

Base7.cpp

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+/*
+Given an integer num, return a string of its base 7 representation.
+
+Example 1:
+Input: num = 100
+Output: "202"
+
+Example 2:
+Input: num = -7
+Output: "-10"
+*/
+
+class Solution {
+public:
+    string s="";
+    string convertToBase7(int num, bool flag=0) {
+       if(num==0){
+           if(s.size()==0) return "0";
+           else reverse(s.begin(),s.end());
+        return flag ? "-"+s : s;
+       }
+       if(num<0) flag=1;
+       s.push_back(abs(num)%7+'0');
+    return convertToBase7(abs(num)/7,flag);
+    }
+};

BaseConversion.cpp

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+/*
+Given four numbers convert:
+
+a, a decimal number to the binary equivalent
+b, a binary to decimal equivalent
+c, decimal to hexadecimal equivalent
+d, hexadecimal to decimal equivalent
+
+Example 1:
+Input :
+a = 6
+b = 110
+c = 20
+d = 2A
+Output:
+110, 6, 14, 42
+Explanation:
+(6)10 = (110)2
+(110)2 = (6)10
+(20)10 = (14)16
+(2A)16 = (42)10
+
+Example 2:
+Input:
+a = 10
+b = 111
+c = 25
+d = FA
+Output:
+1010, 7, 19, 250
+Explanation:
+(10)10 = (1010)2
+(111)2 = (7)10
+(25)10 = (19)16
+(FA)16 = (250)10
+
+Your Task:
+You don't need to read input or print anything. Your task is to complete the function convert() which takes three integers a, b, c, and string d as input parameters and returns the converted numbers as an array of four strings.
+
+Expected Time Complexity: O(log(max(a,b,c)) + |d|)
+Expected Auxiliary Space: O(1)
+*/
+
+class Solution{
+public:
+
+    vector<string> convert(int a,int b,int c,string d)
+    {
+        vector<string> result;
+        string res="";
+        while(a!=0){
+            int rem=a%2;
+            res+=to_string(rem);
+            a/=2;
+        }
+        reverse(res.begin(),res.end());
+        result.push_back(res);
+
+        int i=1, res2=0;
+        while(b!=0){
+            int rem=b%10;
+            res2=res2+rem*i;
+            i*=2;
+            b/=10;
+        }
+        result.push_back(to_string(res2));
+
+        string res3="";
+        while(c!=0){
+            int rem=c%16;
+            if(rem<10){
+                res3+=to_string(rem);
+            }else{
+                res3+=(char)('A'+(c%16)-10);
+            }
+            c/=16;
+        }
+        reverse(res3.begin(),res3.end());
+        result.push_back(res3);
+
+        int res4=0, base=1;
+         for(int i=d.size()-1;i>=0;i--){
+            if(d[i]>=0 and d[i]<='9'){
+                res4+=(d[i]-48)*base;
+                base=base*16;
+            }
+            else if(d[i]>='A' and d[i]<='F'){
+                res4+=(d[i]-55)*base;
+                base*=16;
+            }
+
+        }
+        result.push_back(to_string(res4));
+        return result;
+    }
+};

ConvertToBase-2.cpp

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+/*
+Given an integer n, return a binary string representing its representation in base -2.
+Note that the returned string should not have leading zeros unless the string is "0".
+
+Example 1:
+Input: n = 2
+Output: "110"
+Explantion: (-2)2 + (-2)1 = 2
+
+Example 2:
+Input: n = 3
+Output: "111"
+Explantion: (-2)2 + (-2)1 + (-2)0 = 3
+
+Example 3:
+Input: n = 4
+Output: "100"
+Explantion: (-2)2 = 4
+*/
+
+class Solution {
+public:
+    string baseNeg2(int n) {
+        if(n==0) return "0";
+        string s="";
+        int counter=0;
+        while(n){
+            if(n%2==0){
+                s+='0';
+            }else{
+                s+='1';
+                if(counter%2==1){
+                    n++;
+                }
+            }
+            counter+=1;
+        n=n>>1;
+        }
+        reverse(s.begin(),s.end());
+    return s;
+    }
+};

LargestPrimeFactor.cpp

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+/*
+Given a number N, the task is to find the largest prime factor of that number.
+
+Example 1:
+Input:
+N = 5
+Output:
+5
+Explanation:
+5 has 1 prime factor
+i.e 5 only.
+
+Example 2:
+Input:
+N = 24
+Output:
+3
+Explanation:
+24 has 2 prime factors
+3 and 2 in which 3 is
+greater
+
+Your Task:
+You don't need to read input or print anything. Your task is to complete the function largestPrimeFactor() which takes an integer N as input parameters and returns an integer, largest prime factor of N.
+
+Expected Time Complexity: O(sqrt(N))
+Expected Space Complexity: O(1)
+*/
+
+class Solution{
+public:
+
+    long long int largestPrimeFactor(int N){
+        int result=2;
+        for(int i=2;i*i<=N;i++){
+            while(N%i==0){
+                result=max(result,i);
+                N/=i;
+            }
+        }
+        result=max(result,N);
+    return result;
+    }
+};

ModularExponentiation.cpp

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+/*
+Implement pow(x, n) % M.
+In other words, given x, n and M, find (xn) % M.
+
+Example 1:
+Input:
+x = 3, n = 2, m = 4
+Output:
+1
+Explanation:
+32 = 9. 9 % 4 = 1.
+
+Example 2:
+Input:
+x = 2, n = 6, m = 10
+Output:
+4
+Explanation:
+26 = 64. 64 % 10 = 4.
+
+Your Task:
+You don't need to read or print anything. Your task is to complete the function PowMod() which takes integers x, n and M as input parameters and returns xn % M.
+
+Expected Time Complexity: O(log(n))
+Expected Space Complexity: O(1)
+*/
+
+/// Modular Exponentiation for large numbers
+
+class Solution
+{
+	public:
+		long long int PowMod(long long int x,long long int n,long long int M)
+		{
+		    if(x==0) return 0;
+		    if(n==0) return 1;
+		    if(n==1) return x%M;
+		    long long int y;
+		    if(n%2==0){
+		        y=PowMod(x,n/2,M)%M;
+		        y=(y*y)%M;
+		    }else{
+		        y=x%M;
+		        y=(y*PowMod(x,n-1,M)%M)%M;
+		    }
+		return (long long int)(y+M)%M;
+		}
+};