Merge pull request #89 from ishitakapoor26/Greatest

Largest element in an array

Armstrong.c

@@ -0,0 +1,41 @@
+#include <math.h>
+#include <stdio.h>
+int main() {
+   int low, high, number, originalNumber, rem, count = 0;
+   double result = 0.0;
+   printf("Enter two numbers(intervals): ");
+   scanf("%d %d", &low, &high);
+   printf("Armstrong numbers between %d and %d are: ", low, high);
+
+   // iterate number from (low + 1) to (high - 1)
+   // In each iteration, check if number is Armstrong
+   for (number = low + 1; number < high; ++number) {
+      originalNumber = number;
+
+      // number of digits calculation
+      while (originalNumber != 0) {
+         originalNumber /= 10;
+         ++count;
+      }
+
+      originalNumber = number;
+
+      // result contains sum of nth power of individual digits
+      while (originalNumber != 0) {
+         rem = originalNumber % 10;
+         result += pow(rem, count);
+         originalNumber /= 10;
+      }
+
+      // check if number is equal to the sum of nth power of individual digits
+      if ((int)result == number) {
+         printf("%d ", number);
+      }
+
+      // resetting the values
+      count = 0;
+      result = 0;
+   }
+
+   return 0;
+}

CountingVowelsNC.cpp

@@ -0,0 +1,29 @@
+#include <stdio.h>
+int main()
+{
+    char str[100];
+    char *p;
+    int  vCount=0,cCount=0;
+
+    printf("Enter any string: ");
+    fgets(str, 100, stdin);
+
+    //assign base address of char array to pointer
+    p=str;
+
+    //'\0' signifies end of the string
+    while(*p!='\0')
+    {
+        if(*p=='A' ||*p=='E' ||*p=='I' ||*p=='O' ||*p=='U'
+        		||*p=='a' ||*p=='e' ||*p=='i' ||*p=='o' ||*p=='u')
+            vCount++;
+        else
+            cCount++;
+        //increase the pointer, to point next character
+        p++;
+    }
+
+    printf("Number of Vowels in String: %d\n",vCount);
+    printf("Number of Consonants in String: %d",cCount);
+    return 0;
+}

GreatestElement.c

@@ -0,0 +1,27 @@
+// C program to find maximum in arr[] of size n
+#include <stdio.h>
+
+// C function to find maximum in arr[] of size n
+int largest(int arr[], int n)
+{
+	int i;
+
+	// Initialize maximum element
+	int max = arr[0];
+
+	// Traverse array elements from second and
+	// compare every element with current max
+	for (i = 1; i < n; i++)
+		if (arr[i] > max)
+			max = arr[i];
+
+	return max;
+}
+
+int main()
+{
+	int arr[] = {10, 324, 45, 90, 9808};
+	int n = sizeof(arr)/sizeof(arr[0]);
+	printf("Largest in given array is %d", largest(arr, n));
+	return 0;
+}

MaxMinMatrix.c

@@ -0,0 +1,56 @@
+/* structure is used to return two values from minMax() */
+#include<stdio.h>
+struct pair
+{
+int min;
+int max;
+};
+
+struct pair getMinMax(int arr[], int n)
+{
+struct pair minmax;	
+int i;
+
+/*If there is only one element then return it as min and max both*/
+if (n == 1)
+{
+	minmax.max = arr[0];
+	minmax.min = arr[0];	
+	return minmax;
+}
+
+/* If there are more than one elements, then initialize min
+	and max*/
+if (arr[0] > arr[1])
+{
+	minmax.max = arr[0];
+	minmax.min = arr[1];
+}
+else
+{
+	minmax.max = arr[1];
+	minmax.min = arr[0];
+}
+
+for (i = 2; i<n; i++)
+{
+	if (arr[i] > minmax.max)	
+	minmax.max = arr[i];
+
+	else if (arr[i] < minmax.min)	
+	minmax.min = arr[i];
+}
+
+return minmax;
+}
+
+/* Driver program to test above function */
+int main()
+{
+int arr[] = {1000, 11, 445, 1, 330, 3000};
+int arr_size = 6;
+struct pair minmax = getMinMax (arr, arr_size);
+printf("nMinimum element is %d", minmax.min);
+printf("nMaximum element is %d", minmax.max);
+getchar();
+}

Median.c

@@ -0,0 +1,68 @@
+// A Simple Merge based O(n) solution to find
+// median of two sorted arrays
+#include <stdio.h>
+
+/* This function returns median of ar1[] and ar2[].
+Assumption in this function:
+Both ar1[] and ar2[] are sorted arrays */
+int getMedian(int ar1[], int ar2[], int n, int m)
+{
+	int i = 0; /* Current index of input array ar1[] */
+	int j = 0; /* Current index of input array ar2[] */
+	int count;
+	int m1 = -1, m2 = -1;
+
+	// Since there are (n+m) elements,
+	// There are following two cases
+	// if n+m is odd then the middle
+	//index is median i.e. (m+n)/2
+	if((m + n) % 2 == 1) {
+		for (count = 0; count <= (n + m)/2; count++) {
+			if(i != n && j != m){
+			m1 = (ar1[i] > ar2[j]) ? ar2[j++] : ar1[i++];
+			}
+			else if(i < n){
+			m1 = ar1[i++];
+			}
+			// for case when j<m,
+			else{
+			m1 = ar2[j++];
+			}
+		}
+		return m1;
+	}
+
+	// median will be average of elements
+	// at index ((m+n)/2 - 1) and (m+n)/2
+	// in the array obtained after merging ar1 and ar2
+	else {
+		for (count = 0; count <= (n + m)/2; count++) {
+			m2 = m1;
+			if(i != n && j != m){
+			m1 = (ar1[i] > ar2[j]) ? ar2[j++] : ar1[i++];
+			}
+			else if(i < n){
+			m1 = ar1[i++];
+			}
+			// for case when j<m,
+			else{
+			m1 = ar1[j++];
+			}
+		}
+		return (m1 + m2)/2;
+	}
+}
+
+/* Driver program to test above function */
+int main()
+{
+	int ar1[] = {900};
+	int ar2[] = {5, 8, 10, 20};
+
+	int n1 = sizeof(ar1)/sizeof(ar1[0]);
+	int n2 = sizeof(ar2)/sizeof(ar2[0]);
+	printf("%d", getMedian(ar1, ar2, n1, n2));
+	getchar();
+	return 0;
+}
+

TOH.c

@@ -0,0 +1,12 @@
+#include<stdio.h>
+void TOH(int n,char x,char y,char z) {
+   if(n>0) {
+      TOH(n-1,x,z,y);
+      printf("\n%c to %c",x,y);
+      TOH(n-1,z,y,x);
+   }
+}
+int main() {
+   int n=3;
+   TOH(n,'A','B','C');
+}