**5.21 (Display numbers in a pyramid pattern) Write a nested for loop that prints the following output:

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Sat Oct 24 2020 17:58:24 GMT+0000 (UTC)

Saved by @mahmoud hussein #c++

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{
	for (int i = 0; i < 8; i++)
	{
		for (int k = 0; k <= i; k++)
		{
			cout << "  ";
		}
		for (int j = i; j < 8; j++)
		{
			cout <<" "<< static_cast<char>(j+97-i);

		}
		for (int j = 7; j >= i; j--)
		{
			cout << " " << static_cast<char>(j + 97 - i);
		}
			cout << "\n";
	}
		

}

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c++ loops

@mahmoud hussein

(Occurrence of max numbers) Write a program that reads integers, finds the largest of them, and counts its occurrences. Assume that the input ends with number 0. Suppose that you entered 3 5 2 5 5 5 0; the program finds that the largest is 5 and the occurrence count for 5 is 4 (Fibonacci Series) The Fibonacci series is a series that begins with 0 and 1 and has the property that each succeeding term is the sum of the two preceding terms. For example, the third Fibonacci number is 1 which is sum of 0 and 1. The next is 2, which is a sum of 1 + 1. Write a program that displays the first ten numbers in a Fibonacci series **5.21 (Display numbers in a pyramid pattern) Write a nested for loop that prints the following output: (Compute the greatest common divisor) Another solution for Listing 5.10 to find the greatest common divisor (GCD) of two integers n1 and n2 is as follows: First find d to be the minimum of n1 and n2, then check whether d, d-1, d-2, . . . , 2, or 1 is a divisor for both n1 and n2 in this order. The first such common divisor is the greatest common divisor for n1 and n2. Write a program that prompts the user to enter two positive integers and displays the GCD. sorting

#include<iostream>
using namespace std;
bool issafe(int** arr,int x,int y,int n){
    for(int row=0;row<=x;row++){
        if(arr[x][y]==1){
            return false;
        }
    }
    int row=x;
    int col=y;
    while(row>=0 && col>=0){
        if(arr[row][col]==1){
            return false;
        }
        row--;
        col--;
    }
     row=x;
     col=y;
    while(row>=0 && col<n){
        if(arr[row][col]==1){
            return false;
        }
        row--;
        col++;
    }
    return true ;
}
bool nqueen(int** arr,int x,int n){
    if(x>=n){
        return true;
    }
    for(int col=0;col<=n;col++){
        if(issafe(arr,x,col,n)){
            arr[x][col]==1;
            if(nqueen(arr,x+1,n)){
                return true;
            }
            arr[x][col]=0;  //backtracking
            
        }
    }
    return false;
}
int main(){
    int n;
    cin>>n;
    int** arr=new int*[n];
    for(int i=0;i<n;i++){
        arr[i]=new int[n];
        for(int j=0;j<n;j++){
            arr[i][j]=0;

        }
    }
    if(nqueen(arr,0,n)){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cout<<arr[i][j]<<"";
            }cout<<endl;
        }
    }


    return 0;
}

**8.11 (Game: nine heads and tails)

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;




void decToBinary(int n)
{
    // array to store binary number 
    int binaryNum[3][3];
    
    //converting to binary 
    for (int i = 0; i < 3; i++) 
    {
        for (int j = 0; j < 3; j++)
        {
            binaryNum[i][j] = n % 2;
            n = n / 2;

        }
     }       

    // printing binary> array in reverse order 
    for (int i = 3-1; i >= 0; i--){
        for (int j = 3 - 1; j >= 0; j--)
        {
            if (binaryNum[i][j] == 0)
                cout << "H" << " ";
            else
                cout << "T" << " ";
        }
        cout << endl;
    }
    
 }

int main()
{
    int n;
    cout << "Enter a decimal number between 1 and 512 ";
    cin >> n;

    decToBinary(n);
    return 0;
}

#include<iostream>
using namespace std;

int sum(int arr[], int a);

int main() {
    int arr[1000], n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        cout << sum(arr, n);
    }
}
int sum(int arr[], int a) {
    int i;
    sum = 0;
    for (i=0; i<n; ++i) {
        sum += arr[i];
    }
    return sum;
}#include<iostream>
using namespace std;

int sum(int arr[], int a);

int main() {
    int arr[1000], n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        cout << sum(arr, n);
    }
}
int sum(int arr[], int a) {
    int i;
    sum = 0;
    for (i=0; i<n; ++i) {
        sum += arr[i];
    }
    return sum;
}

(Display the Sundays in a month) Write a program that prompts the user to enter the month and first day of the month, and displays all the Sundays in that month. For example, if the user entered the month 7 for July, and the first day 4 for Wednesday, your program should display the following output: The first day of this month is Wednesday! ... Next Sunday of this month is on 26

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{
	cout << "enter the month ";
	int month;
	cin >> month;
	cout << "Enter The First Day Of The Month ";
	int firstDay;
	cin >> firstDay;
	int nextSunday = 1;
	
		switch (firstDay)
		{
			
		case 1: nextSunday += 7;
			cout << "The first day of this month is sunday "<<endl;
			while (nextSunday < 30)
			{
					cout << "Next Sunday of this month is on " << nextSunday << endl;
									nextSunday += 7 ;
			}
			break;

		case 2: nextSunday += 6;
			cout << "The first day of this month is monday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;
		case 3: nextSunday += 5;
			cout << "The first day of this month is tuesday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;
		case 4: nextSunday += 4;
			cout << "The first day of this month is wednesday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;
		case 5: nextSunday += 3;
			cout << "The first dayof this month is thursday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;
		case 6: nextSunday += 2;
			cout << "The first day of this month is friday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;
		case 7: nextSunday += 1;
			cout << "The first day of this month is saturday " << endl;
			while (nextSunday < 30)
			{
				cout << "Next Sunday of this month is on " << nextSunday << endl;
				nextSunday += 7;
			}
			break;

		}
		
}

#c++ #algorithms #interviewquestions #arrays

Kadane’s Algorithm

int maxSubArraySum(int a[], int size) 
{ 
int max_so_far = 0, max_ending_here = 0; 
for (int i = 0; i < size; i++) 
{ 
	max_ending_here = max_ending_here + a[i]; 
	if (max_ending_here < 0) 
		max_ending_here = 0; 

	/* Do not compare for all elements. Compare only 
		when max_ending_here > 0 */
	else if (max_so_far < max_ending_here) 
		max_so_far = max_ending_here; 
} 
return max_so_far; 
}

Kth element of Spiral Traversal

	public:
	int findK(vector<vector<int>> &a, int n, int m, int k)
    {
        // 1 2 3 4 5 
        // 6 7 8 9 10
        //11 12 13 14 15
        //16 17 18 19 20
        //21 22 23 24 25
        // k=4
        // 1 2 3
        // 4 5 6
        // 7 8 9
        // Your code goes here
        // if(k==1) return a[0][0];
        int lrow=0;
        int lcol=0;
        int urow=n-1;
        int ucol=m-1;
        int count=0;
        int i=0,j=0;
        while(count!=k){
            //iterating from left to right
            for(j=lcol;j<=ucol;j++){
                // cout<<"i:"<<i<<"j:"<<j<<endl;
                count++;
                if(count==k) {
                    goto x;
                }
            }
            ucol--;
            j--;
            //iterating from up to down
            for(i=lrow+1;i<=urow;i++){
                // cout<<"i:"<<i<<"j:"<<j<<endl;
                count++;
                if(count==k) {
                    goto x;
                }
                
            }
            urow--;
            i--;
            //iterating from right to left
            for(j=ucol;j>=lcol;j--){
                // cout<<"i:"<<i<<"j:"<<j<<endl;
                count++;
                if(count==k) {
                    goto x;
                }
                
            }
            lcol++;
            j++;
            //iterating from down to up
             for(i=urow;i>=lrow+1;i--){
                //  cout<<"i:"<<i<<"j:"<<j<<endl;
                count++;
                if(count==k) {
            // cout<<"lrow:"<<lrow<<" urow:"<<urow<<" lcol:"<<lcol<<" ucol:"<<ucol<<endl;
                    goto x;
                }
                
            }
            lrow++;
            i++;
            
        }
        x:
        return a[i][j];
        
    }

566. Reshape the Matrix (Leetcode)

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& mat, int r, int c) {
        vector<vector<int>>ans(r,vector<int>(c));
        int n = mat.size(), m = mat[0].size();
        if(n*m != r*c)  return mat;
        int p = 0, q = 0;
        for(int i = 0; i < r; ++i) {
            for(int j = 0; j < c; ++j) {
                ans[i][j] = mat[p][q];
                q = (q + 1) % m;
                if(q == 0)  p = p + 1;
            }
        }
        return ans;
    }
};

121. Best Time to Buy and Sell Stock (Leetcode)

// TC - O(n) || SC - O(1)
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size() - 1, ans = 0, minSoFar = INT_MAX;
        for(int i = 0; i < n; ++i)  {
            minSoFar = min(minSoFar, prices[i]);
            ans = max(ans, (prices[i] - minSoFar));
        }
        return ans;
    }
};

// we can use auxilary array and store maxSoFar value

350. Intersection of Two Arrays II (Leetcode)

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int,int>mp;
        vector<int> ans;
        for(int i = 0; i < nums2.size(); ++i)   mp[nums2[i]]++;
    
        for(int i = 0; i < nums1.size(); ++i)   { 
            if(mp[nums1[i]] > 0)    {
                mp[nums1[i]]--;
                ans.push_back(nums1[i]);
            }
        }
        return ans;
    }
};

167. Two Sum II - Input array is sorted (Leetcode)

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int i = 0, j = numbers.size() - 1;
        while(i < j) {
            int sum = numbers[i] + numbers[j];
            if(sum == target) return {i + 1, j + 1};
            else if(sum > target)  j--;
            else    i++;
        }
        return {};
    }
};

283. Move Zeroes (Leetcode)

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        for(int i = 0, j = 0; i < nums.size(); i++) {
            if(nums[i]) {
                swap(nums[i],nums[j++]);
            }
        }
    }
};


//2nd approach
class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int zero = 0, one = 0, n = nums.size()-1;
        while(zero < n && one < n)  {
            while(zero < n && nums[zero] != 0)  {
                zero++;
            }
            one = zero;
            while(one < n && nums[one] == 0)    {
                one++;
            }
            swap(nums[zero],nums[one]);
            zero++; 

        }
    }
};

189. Rotate Array (Leetcode)

class Solution {
public:
    void myReverse(vector<int> &nums,int start, int end)   {
        while(start < end)  {
            swap(nums[start++], nums[end--]);
        }
    }
    void rotate(vector<int>& nums, int k) {
        if(nums.size()<2)   return;
        k = k % nums.size();
        myReverse(nums, 0, nums.size() - k - 1);
        myReverse(nums, nums.size()-k, nums.size() - 1);
        myReverse(nums, 0 , nums.size() - 1);
    }
};

977. Squares of a Sorted Array (Leetcode)

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int>ans(nums.size());
        int first = 0, second = nums.size() - 1;
        for(int ind = nums.size() - 1; ind >= 0; ind--)  {
            if(abs(nums[first]) < nums[second]) 
                ans[ind] = nums[second] * nums[second--];
            else    ans[ind] = nums[first] * nums[first++];
        }
        return ans;
        
    }
}


//2nd approch
class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int>ans;
        int first = 0,second = 0;
        while(first<nums.size() && nums[first]<0  ) {
            first++;
        }
        second = first;
        first -= 1;
        while(first>=0 && second<nums.size())   {
            if(abs(nums[first] * nums[first]) < (nums[second] * nums[second]))  {
                ans.push_back(nums[first] * nums[first]);
                first--;
            }
            else {
                ans.push_back(nums[second] * nums[second]);
                second++;
            }
        }
        while(first>=0) {
            ans.push_back(nums[first] * nums[first]);
            first--;
        }
        while(second<nums.size())   {
            ans.push_back(nums[second] * nums[second]);
            second++;
        }
        return ans;

    }
};

217. Contains Duplicate (Leetcode)

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int>st(nums.begin(), nums.end());
        cout<<nums.size()<<"  "<<st.size()<<endl;
        return (nums.size() == st.size()) ? false : true; 
    }
};

88. Merge Sorted Array (Leetcode)

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int p = m-1, q = n-1, ind = nums1.size()-1;
        while(p>=0 && q>=0) {
            if(nums2[q] > nums1[p]) {
                nums1[ind--] = nums2[q];
                q--;
            }
            else    {
                nums1[ind--] = nums1[p];
                p--;
            }
        }
        while(q >= 0)    {
            nums1[ind--] = nums2[q--];
        }
    }
};

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