Merge branch 'q5' of tarfeef101/cs341a1 into master

q5/a1q5.cpp

@@ -0,0 +1,63 @@
+#include <iostream>
+#include <algorithm>
+#include <cstdlib>
+#include <string>
+#include <sstream>
+using namespace std;
+
+int main()
+{
+    string en;
+    int n;
+    getline(cin, en);
+    stringstream cast(en);
+    cast >> n; // get the length of the array we are manipulating
+
+    int * a = (int *) malloc(n * sizeof(int));
+    int * b = (int *) malloc(n * sizeof(int));
+    
+    for (int i = 0; i < n; ++i)
+    {
+        // get each line which is 2 ints, put into A and B
+        string input;
+        getline(cin, input);
+        stringstream nums(input);
+       
+        nums >> a[i];
+	nums >> b[i];
+    }
+    
+    int clen = (n * (n + 1)) / 2; // total possible sums in an array of length n
+    int * c = (int *) malloc(clen * sizeof(int));
+    int tempcount = 0;
+    
+    // Store within c the possible sums inside b
+    for (int i = 0; i < n; ++i)
+    {
+        for (int j = i; j < n; ++j, ++tempcount)
+        {
+            c[tempcount] = b[i] + b[j];
+        }
+    }
+    
+    // sort c, with time complexity n squared log (n squared), which reduces to n squared log n
+    sort(c, c + clen);
+    
+    // for reach sum in a, see if it is in c. n squared checks of log (n squared), or n squared log n
+    for (int i = 0; i < clen; ++i)
+    {
+        for (int j = i; j < n; ++j)
+        {
+            if (binary_search(c, (c + clen), (a[i] + a[j])))
+            {
+                free(a); free (b); free(c);
+		cout << "TRUE" << endl;
+                return 0;
+            }
+        }
+    }
+    
+    free(a); free(b); free(c);
+    cout << "FALSE" << endl;
+    return 0;
+}

q5/q5.cpp

@@ -0,0 +1,100 @@
+#include <iostream>
+#include <cstdlib>
+#include <string>
+
+/*int n;
+int twon = 2 * n;
+int * a;
+int * b;
+int * sums = malloc(2 * sizeof(b));
+int pos = 0;
+
+for (int i = 0; i < twon; i++)
+{
+    for (int j = (i + 1); j < twon; j++, pos++)
+    {
+        sums[pos] = b[i] + b[j];
+    }
+}
+
+sort c;
+
+for (int i = 0; i < twon; i++)
+{
+    for (int j = (i + 1); j < twon; j++, pos++)
+    {
+        if sums.search(a[i] + a[j]) return true;
+    }
+}
+
+return false;
+
+================*/
+
+int n;
+std::cin >> n; // get the length of the array we are manipulating
+
+int * a = malloc(n * sizeof(int));
+int * b = malloc(n * sizeof(int));
+
+for (int i = 0; i < n; ++i)
+{
+    // get each line which is 2 ints, put into A and B
+    std::string input;
+    std::getline(std::cin, input);
+    std::stringstream nums(input);
+    
+    nums >> a[i];
+    nums >> b[i];
+}
+
+/*
+int clen = 2 * n;
+int * C = malloc(clen * sizeof(int)); // to hold the list of possible sums
+int tempcount = 0;
+
+for (int i = 0; i < n; i++)
+{
+    for (int j = (i + 1); j < n; ++j, ++tempcount)
+    {
+        C[tempcount] = (B[i] + B[j]);
+        C[clen - tempcount - 1] = (A[i] + A[j]);
+    }
+} // n squared
+
+sort(C, C + clen); // n^2 log n squared == n^2 log n */
+
+int clen = (n * (n - 1)) / 2; // total possible sums in an array of length n
+int * c = malloc(clen * sizeof(int));
+int tempcount = 0;
+
+// Store within c the possible sums inside b
+for (int i = 0; i < n; ++i)
+{
+    for (int j = (i + 1); j < n; ++j, ++tempcount)
+    {
+        c[tempcount] = b[i] + b[j];
+    }
+}
+
+// sort c, with time complexity n squared log (n squared), which reduces to n squared log n
+std::sort(c, c + clen);
+
+// for reach sum in a, see if it is in c. n squared checks of log (n squared), or n squared log n
+for (int i = 0; i < clen; ++i)
+{
+    for (int j = (i + 1); j < n; ++j)
+    {
+        if (std::binary_search(c, c + clen), (a[i] + a[j]))
+        {
+            free(a); free (b); free(c);
+            cout << 1 << endl;
+            return true;
+        }
+    }
+}
+
+free(a); free(b); free(c);
+cout << 0 << endl;
+return false;
+