Find the first occurence of number in the sorted array

Given an array of numbers, search for the first occurrence of the number

This is relatively easy algorithm. But a bit tricky at the same time.
It can be done with log(n) with almost o(1) space complexity


#include 

/*
 * Find the first occurrence of the given number in index.
 * @author http://gullele.com
 */
void main()
{
    int nums[] = {1,1,3,5,8,8,8,10,12,23,23,55};
    int len = sizeof(nums)/sizeof(int);
    int start = 0, end = len-1, search = 8;
    int found_right = -1;
    while ((end - start) > 1)
    {
        if (found_right != -1)
        {
            if (nums[end] != search)
            {
                break;
            }
            found_right = end;
            end -= 1;
            continue;
        }
        int mid = (end + start)/2;
        int val = nums[mid];
        if (val == search)
        {
            end = mid-1;
            found_right = mid;
        }
        else if(val < search)
        {
            start = mid;
        }
        else
        {
            end = mid;
        }
    }
    if (found_right != -1)
    {
        printf("found at %d", found_right);
    }
    else
    {
        printf("found No where");
    }
}

Changing decimal number to its binary equivalent

Here is bit level solution to change binary equivalent

#include 

/*
 * Printing binary version representation of number
 *
 * Kaleb Woldeareagy <contactkaleb@gmail.com>
 */

void to_binary(unsigned int x);

void main()
{
    unsigned int y=17;
    to_binary(y);
}

void to_binary(unsigned int x)
{
    if (!x|00)
    {
        printf("0");
    }
    else
    {
        int i=0;
        int store[8]; //stack could be used here best
        while (x!=0)
        {
            store[i++]=x&01;
            x>>=1;
        }

        while (i>0)
        {
            printf("%i", store[--i]);
        }
    }
}

Array reversal in Recurrsion

This problem can be performed in o(n) with normal iterative approach.
Here is vanilla flavor of its implementation in recursion

#include 

/**
 * Recursion version of array swap
 *
 * Kaleb Woldearegay
 */

void reverse(int*arr, int i, int size);
void print(int*arr, int size);

void main()
{
    int nums[] = {4,5,2,13,0}, i=0, size=5;
    i = 0;
    print(nums, 5);
    reverse(nums, i, size-1);
    print(nums, 5);
}
void reverse(int*stk, int i, int size)
{
    if (i>=size)
    {
        return;
    }
    else
    {
        int tmp = stk[i];
        reverse(stk, i+1, size-1);
        stk[i] = stk[size];
        stk[size] = tmp;
    }
}

void print(int*nums, int size)
{
    int i=0;
    while (i<size)
    {
        printf("%in", nums[i]);
        i++;
    }
}

Implementing tokenizer and adding tokens to the linked list

Here we go

#include 
#include 
#include 
struct Node
{
    char * value;
    struct Node *next;
};
struct Node *head = NULL;
struct Node *last = NULL;
char* substr(char*str, int start, int length)
{
    const char* from = str;
    char *to = (char*) malloc(sizeof(str)/sizeof(char));
    strncpy(to, from+start, length);
    return to;
}

void addToLinkedList(char* value)
{
    struct Node *curr_node = malloc(sizeof(struct Node));
    char * newtemp = malloc(strlen(value));
    strncpy(newtemp, value, strlen(value));
    curr_node->value = newtemp;
    curr_node->next = NULL;
    if (head == NULL)
    {
        head = curr_node;
        last = head;
    }
    else
    {
        last->next = curr_node;
        last = curr_node;
    }

}

void printLinkedList()
{
    struct Node * curr_node = head;
    while (curr_node !=NULL)
    {
        printf("%sn", curr_node->value);
        curr_node = curr_node->next;
    }
}
void main()
{
    int delimiter_size = 3, i=0, j=0;
    char* delimit = "abc";
    char* text = "someabccodingabctoabcparseab";
    int length = strlen(text);
    char* temp = (char*)malloc(length);
    while(i < length)
    {
        if (text[i] == delimit[0]) //check if the begin
        {
            //check if the 
            if (strcmp(substr(text, i, delimiter_size), delimit) == 0)
            {
                addToLinkedList(temp);
                i+=delimiter_size;
                j=0;
                memset(temp, '', strlen(temp));
                continue;
            }
        }
        temp[j] = text[i];
        i++;
        j++;
    }
    if (strlen(temp))
    {
        addToLinkedList(temp);
    }
    printLinkedList();
}

Kadane’s algorithm in C – Dynamic Programming

How do we find a keys that hold maximum consecutive values in sum from the given array?
The usual approach could be O(n^2).

But Kadanes algorithm can perform this same problem in a linear time.
here is the implementation using C

#include <stdio.h>

/*
 * @author http://gullele.com
 * 
 */
void main()
{
        printf("Kadane's Algorithmn");
        int nums[] = {-1,2,3,-9,8,7,2};
        int start_index;
        int end_index;
        int sum = maximum_consequential_sum(nums, &start_index, &end_index);
        printf("maximum sum is %d n",sum );
        printf("maximum index starts at %d n", start_index);
        printf("maximum index ends at %d n", end_index);
}
/*
 * Get the maximum subsequent sum from the given array
 * function performing kadane's Algorithm 
 */
int maximum_consequential_sum(int* nums, int*start_index, int*end_index)
{
        int max_start_index = 0, max_end_index = 0;
        int max_sum = 0;
        int cur_index, cur_max_sum = 0;
        for(cur_index = 0; cur_index <= max_sum ; cur_index++)
        {
                cur_max_sum += nums[cur_index];
                if (cur_max_sum > max_sum)
                {
                        max_sum = cur_max_sum;
                        max_end_index = cur_index;
                }
                if (cur_max_sum < 0)
                {
                        cur_max_sum = 0 ;
                        max_start_index = cur_index + 1;
                }
        }
        *start_index = max_start_index;
        *end_index = max_end_index;
        return max_sum;
}

The above code is, of course, just for illustration purpose. Make sure to check
for empty array and possible division by zero stuff..

To run this on Linux

1. Save the above file as kadanes.c
2. go to terminal and locate to the file
3. gcc kadanes.c -o kadane
4 run as ./kadane

** you can use this code as you like. As the same time, I am not responsible for anything, in case if you use it.
Thanks

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