/*
* 1.Given a sorted array a and a sum k print
* all pairs of indexes i and j such that
* a[i]+a[j]=k.
*/
void printPairsWithSum(int *array, int k, int len) {
int startIndex = 0, endIndex = len - 1, sum;
while (startIndex < endIndex) {
sum = array[startIndex] + array[endIndex];
if (sum == k) {
printf("%d %d \n", startIndex, endIndex);
startIndex++;
} else if (sum > k)
endIndex--;
else
startIndex++;
}
}
int main() {
int *array = malloc(sizeof(int) * 100);
int len, i;
scanf("%d", &len);
for (i = 0; i < len; i++) {
scanf("%d ", &array[i]);
}
printPairsWithSum(array, 12, len);
return 0;
}
My Learnings
Friday, July 19, 2013
Find a pair of elements from an array whose sum equals a given number
Monday, July 15, 2013
Sorting a linked list
/**Algorithm best suited to sort a linked list.
*
* Write a C program to implement that algorithm.
* Modified quick sort should work.
*
* ex: 4 8 2 9 5 3
* 4 2 8 9 5 3
* 4 2 3 9 5 8
*/
#include "Common.h"
typedef struct _node {
struct _node *next;
int value;
} Node;
void swap(Node* node1, Node* node2) {
int temp = node1->value;
node1->value = node2->value;
node2->value = temp;
}
void sortLinkedList(Node *head, Node *lastNode) {
if (head == NULL )
return;
// if last two nodes and already sorted return
if (head == lastNode || head->next == lastNode)
return;
Node* nextNode = head->next;
if (nextNode->next == lastNode)
{
if (head->value < head->next->value)
return;
else
{
swap(head, head->next);
return;
}
}
Node *pivot = head;
Node *startNode = head->next, *endNode = head->next,
*currSwapNode = startNode;
while (1) {
while (startNode != lastNode) {
if (startNode->value < pivot->value) {
currSwapNode = startNode;
startNode = startNode->next;
} else {
break;
}
}
if (startNode == lastNode)
break;
endNode = startNode == lastNode ? startNode : startNode->next;
while (endNode != lastNode && endNode->value >= pivot->value) {
endNode = endNode->next;
}
if (endNode != lastNode && startNode != lastNode)
{
swap(startNode, endNode);
currSwapNode = startNode;
}
else
break;
startNode = startNode->next;
}
if (currSwapNode != NULL && pivot->value > currSwapNode->value) {
swap(pivot, currSwapNode);
sortLinkedList(head, currSwapNode);
sortLinkedList(currSwapNode->next, lastNode);
}
else
sortLinkedList(currSwapNode, lastNode);
}
int main() {
int numNumbers, number, i;
Node *head = NULL, *current = NULL;
printf("Enter the numbers : ");
scanf("%d", &numNumbers);
for (i = 0; i < numNumbers; i++) {
scanf("%d", &number);
Node *node = malloc(sizeof(Node));
node->value = number;
node->next = NULL;
if (current != NULL ) {
current->next = node;
}
if (head == NULL ) {
head = node;
}
current = node;
}
current = head;
while (current != NULL ) {
printf("%d ", current->value);
current = current->next;
}
printf("\n");
sortLinkedList(head, NULL );
current = head;
while (current != NULL ) {
printf("%d ", current->value);
current = current->next;
}
return 0;
}
Sunday, September 23, 2012
Find common elements in given two Binary Search Trees
/*
* Two BST are given find common elements in both
* Algo works like this:
* Iteratively traverse both the tree in inorder fashion
* If the first tree node is less than the second tree move the
* first tree forward else if the second tree node is less than the
* first tree move second tree forward. If equal move both tree forward
* Do this operation till either one of the tree is complete.
*
*/
void CommonElementsInBST(TreeNode* root1, TreeNode* root2) {
Stack *stack1 = createStack();
Stack *stack2 = createStack();
int commonElementCount = 0;
do {
while (root1 != NULL ) {
push(stack1, root1);
root1 = root1->left;
}
while (root2 != NULL ) {
push(stack2, root2);
root2 = root2->left;
}
if (isEmpty(stack1) != 0 && isEmpty(stack2)) {
root1 = (TreeNode*)top(stack1);
root2 = (TreeNode*)top(stack2);
if (*((int*)root1->object) < *((int*)root2->object)) {
pop(stack1);
root1 = root1->right;
} else if (*((int*)root1->object) > *((int*)root2->object)) {
pop(stack2);
root2 = root2->right;
} else {
pop(stack1);
pop(stack2);
printf("%d ", *((int*) root1->object));
root1 = root1->right;
root2 = root2->right;
commonElementCount++;
}
}
} while (((isEmpty(stack1) != 0) || root1 != NULL) && ((isEmpty(stack2) != 0) || root1 != NULL));
printf("Number of common elements are : %d", commonElementCount);
}
Sunday, September 9, 2012
Stack Operations
Stack create, push, pop, empty operations
typedef struct _stack {
int top;
void* array[20];
} Stack;
Stack* createStack()
{
Stack *stack = malloc(sizeof(Stack));
memset(stack, 0, sizeof(Stack));
return stack;
}
void push(Stack* stack, void* elem)
{
stack->array[stack->top] = elem;
stack->top += 1;
}
void* pop(Stack* stack)
{
void* elem = stack->array[stack->top-1];
stack->top -= 1;
return elem;
}
void* top(Stack* stack)
{
if (stack->top == 0)
return NULL;
return stack->array[stack->top-1];
}
int isEmpty(Stack* stack)
{
return stack->top;
}
Inorder Traversal of binary tree
Inorder traversal of binary tree - Recursive
Inorder traversal of binary tree - Iterative
typedef struct _binaryTreeNode
{
struct _binaryTreeNode *left;
struct _binaryTreeNode *right;
void* object;
} TreeNode;
void inOrderTraversalRecursive(TreeNode* root) {
if (root == NULL )
return;
inOrderTraversalRecursive(root->left);
printf("%d ", *(int *) root->object);
inOrderTraversalRecursive(root->right);
}
Inorder traversal of binary tree - Iterative
void inOrderTraversalIterative(TreeNode* root) {
if (root == NULL )
return;
Stack *tempStack = createStack();
do {
while (root != NULL ) {
push(tempStack, root);
root = root->left;
}
if (isEmpty(tempStack) != 0) {
root = pop(tempStack);
printf("%d ", *((int*) root->object));
root = root->right;
}
} while ((isEmpty(tempStack) != 0) || root != NULL );
}
PS: Stack related functions are here.
Mirror a Binary Tree
Following algorithm would create a mirror of a Tree
void mirror(TreeNode* root) {
if (root == NULL )
return;
TreeNode* temp = root->left;
root->left = root->right;
root->right = temp;
mirror(root->left);
mirror(root->right);
}
Tuesday, September 4, 2012
Removing consecutive Zeros
/*Remove all two consecutive zeros in the given string.
Example: a3409jd00dk000d
Output: a3409jddk000d
Note: If there are less/more than two consecutive zeros, they should not be replaced.
*/
void remove2zeros(char *input) { char *temp = input; while (*temp != '\0') { if (*temp == '0' && *(temp - 1) != '0' && *(temp + 1) == '0' && *(temp + 2) != '0') { temp = temp + 2; } else { if (input != temp) { *input++ = *temp++; } else { temp++; input++; } } } *input = '\0'; }
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