C Challenge 3 (Unreviewed)

challenge_3/c/tyleroar/README.md

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+#Majority Element
+
+##Premise
+
+-	Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
+
+-	You may assume that the array is non-empty and the majority element always exist in the array.
+
+for example, given array = [2,2,3,7,5,7,7,7,4,7,2,7,4,5,6,7,7,8,6,7,7,8,10,12,29,30,19,10,7,7,7,7,7,7,7,7,7] your program should return 7
+
+[Testing](https://github.com/YearOfProgramming/2017Challenges/tree/testing#testing)
+
+Expected input:
+
+    2,2,3,7,5,7,7,7,4,7,2,7,4,5,6,7,7,8,6,7,7,8,10,12,29,30,19,10,7,7,7,7,7,7,7,7,7
+
+Expected output:
+
+    7

challenge_3/c/tyleroar/src/challenge3.c

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+#include <stdio.h>
+#include <string.h>
+#include <stdlib.h>
+struct bst {
+    int x;
+    int occ;
+    struct bst *left;
+    struct bst *right;
+};
+void freeBST(struct bst *head) {
+    if (head==NULL) return;
+    if (head->left!=NULL) freeBST(head->left);
+    if (head->right!=NULL) freeBST(head->right);
+    free(head);
+}
+int addNum(int num, struct bst *head) {
+    struct bst *cur = head;
+    while (1) {
+        if (num==cur->x) {
+            cur->occ++;    //if the number is already in the tree increase it's occurence
+            return cur->occ;
+        }
+        if (num<cur->x) {
+            if (cur->left==NULL) {
+                struct bst *new = malloc(sizeof(struct bst));
+                new->x = num;
+                new->occ = 1;
+                new->left = NULL;
+                new->right = NULL;
+                cur->left = new;
+                return new->occ;
+            } else cur = cur->left;
+        } else if (num>cur->x) {
+            if (cur->right ==NULL) {
+                struct bst *new = malloc(sizeof(struct bst));
+                new->x = num;
+                new->left = NULL;
+                new->right = NULL;
+                new->occ=1;
+                cur->right = new;
+                return new->occ;
+            } else cur=cur->right;
+        }
+    }
+}
+char *readLine() {
+    int length = 1;
+    char *string = NULL;
+    char c;
+    while (EOF!=(c=getchar()) && '\r'!=c && '\n' != c) {
+        string=realloc(string,length);
+        string[length-1]=c;
+        length++;
+    }
+    string[length-1]='\0';
+    return string;
+}
+int majorityElement(int *nums, int length) {
+    if (!nums) return -1;
+    int element = nums[0];
+    struct bst *head = malloc(sizeof(struct bst));
+    head->x=nums[0];
+    head->occ =1;
+    head->left = NULL;
+    head->right = NULL;
+    for (int i=1; i<length; i++) {
+        int occ = addNum(nums[i],head);
+        if (occ>(length/2)) element=nums[i];
+    }
+    if (head) freeBST(head);
+    return element;
+}
+int main() {
+    int *nums = malloc(sizeof(int));
+    int length=0;
+    char *input = readLine();
+    if (input==NULL) {
+        printf("unable to read input!\n");
+        return -1;
+    }
+    char *ptr=NULL;
+    ptr = strtok(input,",");
+    while (ptr!=NULL) {
+        int a = atoi(ptr);
+        length++;
+        if (length>1) nums=realloc(nums,(length+1)*sizeof(int));
+        nums[length-1]=a;
+        ptr=strtok(NULL, ",");
+    }
+
+    printf("%d\n", majorityElement(nums, length));
+
+    if (input) free(input);
+    if (nums) free(nums);
+    return 0;
+}