local

Author: mostafa-saad

18-Programming-4kids/10_homework_01_answer.cpp

@@ -0,0 +1,27 @@
+#include<iostream>
+using namespace std;
+
+int main() {
+	int n, q, a[200];
+
+	cin >> n;
+	for (int i = 0; i < n; i++)
+		cin >> a[i];
+
+	cin >> q;
+	while (q--) {
+		int num;
+		cin >> num;
+
+		int pos = -1;
+		// search from the end
+		for (int i = n-1; i >= 0; --i) {
+			if (a[i] == num) {
+				pos = i;
+				break;
+			}
+		}
+		cout << pos << "\n";
+	}
+	return 0;
+}

18-Programming-4kids/10_homework_01_answer_1.cpp

@@ -0,0 +1,28 @@
+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+
+#include<iostream>
+using namespace std;
+
+int main() {
+	// As values are 0-500, we can make array of 501 mark the last value in it
+	// Then we answer the queries directly
+
+	const int N = 500 + 1;
+	int n, q, x, ans[N];
+
+	for (int i = 0; i < N; i++)
+		ans[i] = -1; /** at first the answer of any number is -1 **/
+
+	cin >> n;
+	for (int i = 0; i < n; i++) {
+		cin >> x;
+		ans[x] = i;/** update the answer for x to be i **/
+	}
+	int num;
+	cin >> q;
+	while (q--) {
+		cin >> num;
+		cout << ans[num] << endl;
+	}
+	return 0;
+}

18-Programming-4kids/10_homework_01_answer_2.cpp

@@ -1,35 +1,29 @@
 // By Basel Bairkdar https://www.facebook.com/baselbairkdar
 
-#include<bits/stdc++.h>
-
+#include<iostream>
 using namespace std;
 
-const int N=205;
 
-int n,a[N];
+int main() {
+	const int N = 200;
+	int n, a[N];
+
+	cin >> n;
+	for (int i = 0; i < n; i++) {
+		cin >> a[i];
+	}
 
-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;
+	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;
+			break;
+		}
+	}
+	if (increasing) {
+		cout << "YES\n";
+	} else {
+		cout << "NO\n";
+	}
+	return 0;
 }

18-Programming-4kids/10_homework_02_answer.cpp

@@ -1,42 +1,36 @@
 // By Basel Bairkdar https://www.facebook.com/baselbairkdar
 
-#include<bits/stdc++.h>
-
+#include<iostream>
 using namespace std;
 
-const int N=205;
+int main() {
+
+	const int N = 100;
+
+	int n, a[N], mn = 10000, mx = -1;
 
-int n,a[N],mn=1e9,mx=-1,index_mn,index_mx;
+	cin >> n;
+	for (int i = 0; i < n; i++) {
+		cin >> a[i];
+		if (a[i] < mn) {
+			mn = a[i];
+		}
+		if (a[i] > mx) {
+			mx = a[i];
+		}
+	}
 
-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
+	for (int i = 0; i < n; i++) {
+		if (a[i] == mn)
+			a[i] = mx;
+		else if (a[i] == mx)
+			a[i] = mn;
+	}
 
-    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;
+	for (int i = 0; i < n; i++) {
+		if (i)
+			cout << " ";
+		cout << a[i];
+	}
+	return 0;
 }

18-Programming-4kids/10_homework_03_answer.cpp

@@ -0,0 +1,46 @@
+// By Basel Bairkdar https://www.facebook.com/baselbairkdar
+#include<iostream>
+using namespace std;
+
+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.
+	 **/
+	const int N = 200;
+	int n, a[N], mn[3], mx;
+
+	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_04_answer.cpp

@@ -1,36 +1,32 @@
 // By Basel Bairkdar https://www.facebook.com/baselbairkdar
+#include<iostream>
+using namespace std;
 
-#include<bits/stdc++.h>
+int main() {
+	const int N = 200;
 
-using namespace std;
+	int n, a[N];
 
-const int N=205;
+	// let's use some big value and later minimize
+	int mn = 10000000;
 
-int n,a[N],mn=1e9;
+	cin >> n;
+	for (int i = 0; i < n; i++)
+		cin >> a[i];
 
-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;
+	/**
+	 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;
 }