local

Author: mostafa-saad

18-Programming-4kids/10_homework_01_answer.cpp

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+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+#include<bits/stdc++.h>
+
+using namespace std;
+
+const int N=205;
+
+int n,a[N],mn=1e9;
+
+int main()
+{
+    cin >> n;
+    for (int i=0;i<n;i++)
+    {
+        cin >> a[i];
+    }
+    /**
+    let's calculate Ai+Aj+j-i for every pair (i,j)
+    such that i < j
+    this can be done using nested for loops.
+    **/
+    for (int i=0;i<n;i++)
+    {
+        for (int j=i+1;j<n;j++)
+        {
+            int tmp=a[i]+a[j]+j-i;
+            if (tmp<mn)
+            {
+                mn=tmp;
+            }
+        }
+    }
+    cout << mn << endl;
+    return 0;
+}

18-Programming-4kids/10_homework_02_answer.cpp

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+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+
+#include<bits/stdc++.h>
+
+using namespace std;
+
+const int N=205;
+
+int n,a[N];
+
+int main()
+{
+    cin >> n;
+    for (int i=0;i<n;i++)
+    {
+        cin >> a[i];
+    }
+    bool increasing=1; /** we will assume that the array is increasing at first **/
+    for (int i=1;i<n;i++)
+    {
+        if (a[i]<a[i-1])
+        {
+            increasing=0;
+        }
+    }
+    if (increasing)
+    {
+        cout << "YES\n";
+    }
+    else
+    {
+        cout << "NO\n";
+    }
+    return 0;
+}

18-Programming-4kids/10_homework_03_answer.cpp

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+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+
+#include<bits/stdc++.h>
+
+using namespace std;
+
+const int N=205;
+
+int n,a[N],mn=1e9,mx=-1,index_mn,index_mx;
+
+int main()
+{
+    /**
+    store minimum value in mn, maximum value in mx
+    store the index of mn in index_mn, index of mx in index_mx
+
+    why initiating mn with large number and mx with small number ?
+    try mn=0 and see what happens
+    **/
+    cin >> n;
+    for (int i=0;i<n;i++)
+    {
+        cin >> a[i];
+        if (a[i]<mn)
+        {
+            mn=a[i];
+            index_mn=i;
+        }
+        if (a[i]>mx)
+        {
+            mx=a[i];
+            index_mx=i;
+        }
+    }
+    a[index_mn]=mx;
+    a[index_mx]=mn;
+    for (int i=0;i<n;i++)
+    {
+        cout << a[i] << ((i==n-1)?"\n":" ");
+    }
+    return 0;
+}

18-Programming-4kids/10_homework_04_answer.cpp

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+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+
+#include<bits/stdc++.h>
+
+using namespace std;
+
+const int N=205;
+
+int n,a[N],mn[3],mx;
+
+int main()
+{
+    /**
+    we will put the first three elements from the array
+    in a new array called mn which will store the minimum
+    three values so far.
+
+    now we will iterate over the input array and see
+    if the current element is less than the maximum element
+    in array mn, if so we will make change.
+    **/
+    cin >> n;
+    for (int i=0;i<n;i++)
+    {
+        cin >> a[i];
+        if (i<3)
+        {
+            mx=max(mx,a[i]);
+            mn[i]=a[i];
+        }
+    }
+    for (int i=3;i<n;i++)
+    {
+        if (a[i]<mx)
+        {
+            for (int j=0;j<3;j++)
+            {
+                if (mn[j]==mx)
+                {
+                    mn[j]=a[i];
+                    break;
+                }
+            }
+            // update mx value
+            for (int j=0;j<3;j++)
+            {
+                if (mn[j]>mx)
+                {
+                    mx=mn[j];
+                }
+            }
+        }
+    }
+    for (int i=0;i<3;i++)
+    {
+        cout << mn[i] << ((i==n-1)?"\n":" ");
+    }
+    return 0;
+}

18-Programming-4kids/10_homework_05_answer.cpp

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+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+
+#include<bits/stdc++.h>
+
+using namespace std;
+
+const int N=205;
+
+int n,a[N],mn=1e9;
+
+int main()
+{
+    cin >> n;
+    for (int i=0;i<n;i++)
+    {
+        cin >> a[i];
+    }
+    /**
+    let's calculate Ai+Aj+j-i for every pair (i,j)
+    such that i < j
+    this can be done using nested for loops.
+    **/
+    for (int i=0;i<n;i++)
+    {
+        for (int j=i+1;j<n;j++)
+        {
+            int tmp=a[i]+a[j]+j-i;
+            if (tmp<mn)
+            {
+                mn=tmp;
+            }
+        }
+    }
+    cout << mn << endl;
+    return 0;
+}

18-Programming-4kids/10_homework_06_answer.cpp

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+// By Amr Keleg https://www.facebook.com/amr.keleg
+
+#include <iostream>
+using namespace std;
+
+int main() {
+	// The size of the array should be larger than
+	// the maximum value of N
+	int arr[1000];
+	
+	int N;
+	cin>>N;
+	
+	for(int i=0; i<N; i++)
+		cin>>arr[i];
+		
+	// We need to compare the value at the left half
+	// to those in the right half
+	// The value at index (0) should be compared to the value at index (N-1)
+	// Then the value at index (1) should be compared to the value at index (N-2)
+	// So, clearly we need to increment the left index and decrement the right one
+	// On reaching the index N/2, we are sure that we have compared all the elements
+	// of both sides so we don't need to do anything and the array is palindrome
+	for (int i=0; i< N/2; i++){
+		if (arr[i] != arr[N-1-i])
+		{
+			cout<<"NO";
+			return 0;
+		}
+	}
+	cout<<"YES";
+	return 0;
+}

18-Programming-4kids/10_homework_07_answer.cpp

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+// By Amr Fahim https://www.facebook.com/amrfahim2020
+// Homework 7 solution 
+#include <iostream>
+using namespace std ;
+
+// We use 2 frequency arrays one for the positive numbers and the other for the negative 
+
+int n , neg_num[500 +10] , pos_num[270 + 10 ]  , max_num , max_freq ;
+
+int main()
+ {
+    cin >> n ;
+    int arr[n] ;
+    
+    for (int i = 0 ; i < n ;i++ )
+    {
+         cin >> arr[i] ; 
+         // Check the sign of the input value 
+         //to know which array we should add the deal with
+         if (arr[i] >= 0 ) 
+         {
+             pos_num[arr[i]]++ ;
+
+         // Comparing the value we have now with the max value  
+            if (pos_num[arr[i]] > max_freq ) 
+            {
+                max_freq = pos_num[arr[i]] ;
+                max_num = arr[i] ;
+                
+            }  
+
+         }
+         else if (arr[i] < 0 ) 
+         {
+             neg_num[arr[i]]++ ;
+
+              // Comparing the value we have now with the max value  
+            if (neg_num[arr[i]] > max_freq )
+            {
+                max_freq = neg_num[arr[i]] ;
+                max_num = arr[i] ;
+            }
+    }
+
+    
+}
+    cout << max_num <<" has repeated " << max_freq <<" times " ;
+}

18-Programming-4kids/10_homework_08_answer.cpp

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+// By Amr Fahim https://www.facebook.com/amrfahim2020
+// Homework 8 solution
+
+#include <iostream>
+using namespace std ;
+
+int n , x , occurrence[10] ;
+
+int main()
+ {
+    cin >> n ;
+    
+    for (int i = 0 ; i < n ;i++ )
+    {
+         cin >> x ; 
+        
+         while(x)
+         {
+             int digit = x%10 ;
+             occurrence[digit]++ ;
+             x /= 10 ;
+         }
+         
+    }
+    for (int i = 0 ; i <= 9 ; i++)
+    {
+        cout << i <<" " << occurrence[i] << endl ;
+    }
+    
+}
+

18-Programming-4kids/10_homework_09_answer.cpp

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+// By Amr Fahim https://www.facebook.com/amrfahim2020
+// Homework 9 solution
+#include <iostream>
+using namespace std ;
+
+const int N = 2e5+10 ;
+int occurrence[N] , sequence[N] , indx ;
+
+int main()
+ {
+    cin >> indx ;
+    sequence[0] = 0 ;
+    occurrence[0] = 1 ;
+
+    for (int i = 1 ; i <= indx ;i++ )
+    {
+        int cur = sequence[i-1] - (i-1) -1  ;
+
+        if (cur < 0 || occurrence[cur] )
+        {
+            cur = sequence[i-1] + (i-1) + 1 ;
+        }
+         sequence[i] = cur ;
+         occurrence[cur] = 1 ;
+     
+    }
+   
+   cout << sequence[indx] ;
+    
+}
+
+