The maximum-subarray problem

write a program to find the nonempty, contiguous subarray of an Array whose values have the largest sum.This contiguous subarray is called the maximum subarray.
/*
 * Brute Force
 */
int * FindMaxSubArray0(int A[], int n)
{
    int i,j,sum,high,low;
    int *max = (int *)malloc(n*sizeof(int));
    int *end_pos = (int *)malloc(n*sizeof(int));
    int *p = (int *) malloc(3* sizeof(int));
    for(i = 0; i < n; i++) {
          max[i] = A[i];
          end_pos[i] = i;
          sum = max[i];
          for(j = i+1; j < n; j++) {
                sum = sum + A[j];
                if(sum > max[i]) {
                        max[i] = sum;
                        end_pos[i] = j;
                }
          }
    }
    p[0] = 0;
    p[1] = end_pos[0];
    p[2] = max[0];
    for(i = 1; i < n; i++) {
          if(max[i] > p[2]) {
                  p[2] = max[i];
                  p[0] = i;
                  p[1] = end_pos[i];
          }
    }
    return p;
}
/*
 * Divide-and-Conquer strategy
 */
int * FindMaxCrossSubArray(int A[],int low,int high,int mid)
{
     int i,sumleft = A[mid],sumright = A[mid+1], sum,left,right;
     int *p =(int *)malloc(sizeof(int) * 3);
     sum = sumleft;
     left = mid;
     for(i = mid-1; i > low; i--) {
           sum = sum+ A[i];
           if(sum > sumleft) {
                   sumleft = sum;
                   left = i;
           }
     }
     sum=sumright;
     right = mid+1;
     for(i = mid+2; i < high; i++) {
           sum = sum + A[i];
           if(sum > sumright) {
                   sumright = sum;
                   right = i;
           }
     }
     p[0] = left;
     p[1] = right;
     p[2] = sumleft+sumright;
     return p;
}
int * FindMaxSubArray1(int A[], int low, int high)
{
     int mid, *left=NULL, *right=NULL, *cross=NULL, *p = NULL;
     if(low == high) {
            p =(int *)malloc(sizeof(int) * 3);
            p[0] = p[1] = low;
            p[2] = A[low];
            return p;
     } else {
            mid = (low+high)/2;
            left = FindMaxSubArray1(A,low,mid);
            right = FindMaxSubArray1(A,mid+1,high);
            cross = FindMaxCrossSubArray(A,low,high,mid);
            if(left[2] >= right[2] && left[2] >= cross[2]) {
                       return left;
            } else if(right[2] >= left[2] && right[2] >= cross[2]) {
                   return right;
            } else {
                   return cross;
            }
     }
}
/*
 * Dynamic Programmming Methodology
 */
int * FindMaxSubArray2(int A[], int n)
{
    int *max, *end_pos, sum=0,*p,i;
    max = (int*) malloc(sizeof(int)*n);
    end_pos = (int*) malloc(sizeof(int)*n);
    p = (int*) malloc(sizeof(int)*3);
    for(i = 0; i < n; i++) {
          max[i] = A[i];
          end_pos[i] = i;
    }
    for(i = 1; i < n; i++) {
          sum = max[i-1] + max[i];
          if(sum > max[i]) {
                  max[i] = sum;
                  end_pos[i] = i;
          }
    }
    p[2] = max[0];
    for(i = 1; i < n; i++) {
          if(max[i] > p[2]) {
                    p[2] = max[i];
                    p[1] = i;
          }
    }
    for(i = p[1]; i > 0; i--) {
          if(end_pos[i] == i) {
                        p[0] = i;
                        break;
          }
    }
}

Inversion

Let A[1..n] be an array of n distinct numbers.If i < j and A[i] > A[j], then the pair (i,j) is called an inversion of A.
Consider the array (2,3,8,6,1). Inversions in array are (8,6), (2,1), (3,1) , (8,1), (6,1).

Write program to count the number of inversion in array.

/*
 * CountInverion0 : This is modified version of insertion sort. Time Complexity : O(n^2)
 */
int CountInversion0(int a[], int n)
{
    int i,j,count=0;
    for(i=1; i<n; i++) {
             for(j=0; j<i; j++)
                      if(a[j] > a[i])
                              count++;
    }
    return count;
}

/*
 * CountInversion1 : Modifed version of Merge sort. Uses Divide and Conquer Methodlogy.
 * TimeComplexity: O(n*log(n)).
 */
int CountInversion1(int A[], int start, int end)
{
    int i,x,y;
    int z=0;
    if((end-start) == 0) return 0;
    else if(start < end) {
        i = (start+end)/2;
        x = CountInversion1(A, start , i);
        y = CountInversion1(A, i+1, end);
        z = CountSplitInversion(A,start,end,i);
        return x+y+z;
    }
}
/*
 * Helper function: sorts array and also count the number of inversion.
 * Hint : in case of merge sort, while merging if we copy the number from 2nd array to array,
 * number of elements lefts in first array adds to the number of inversions.
 */
int CountSplitInversion(int a[],int start, int end, int div)
{
    int i,j,k,lc = (div-start)+1,rc = (end-div),*l,*r,count = 0;
    l = (int *) malloc(sizeof(int) * (lc+1));
    r = (int *) malloc(sizeof(int) * (rc+1));
    for(i=0; i < lc; i++)
        l[i] = a[start + i];
    for(i=0; i < rc; i++)
        r[i] = a[div+i+1];
    for(k=start,i=0,j=0; k <= end && (i < lc) && (j <rc) ;k++)
    {
        if(l[i] <= r[j]) {
            a[k] = l[i]; i++;
        }
        else {
            a[k] = r[j]; j++;
            count +=  (lc - i);
        }
    }
    if(i < lc) {
        for(;k<=end && i<lc ;k++)
            a[k] = l[i];
    }
    if(j < rc) {
        for(;k<=end && j<rc ;k++)
            a[k] = r[j];
    }
    return count;
}

Write a recursive procedure for Insertion Sort

Do recursion and after that insert the value.
/*
 * Helper function, Insert val at proper position
 */ 
void insert(int a[], int n, int val)
{
     int i;
     for(i=n-1; i >= 0  && a[i] < val; i--)
                a[i+1] = a[i];
     a[i+1] = val;
}
/*
 * Recursive version of Insertion Sort
 */
void insertRsort(int a[], int n)
{
     if(n>0) {
             insertRsort(a, n-1);
             insert(a, n, a[n]);
     }
}

Print a singly-linked list backwards, in linear time

/*
 * Print a singly-linked list backwards in linear time.
 */
void print(nodeptr head)
{
        if(head == NULL) return;
        print(head->next);
        printf("%d  ",head->data);
}

Tree - List Problem

Write a recursive function treeToList(Node root) that takes an ordered binary tree and rearranges the internal pointers to make a circular doubly linked list out of the tree nodes.

/********************************************************
 ****      The Great Tree List Recursion Problem     ****
 *******************************************************/
struct node;

typedef struct node *nodeptr;

struct node {
    nodeptr left;
    int data;
    nodeptr right;
};

/*
 * given two list nodes, join them together so the second
 * immediately follows the first.
 */
void join(nodeptr a, nodeptr b)
{
    a->right = b;
    b->left = a;
}
/*
 * given two circular doubly linked lists, append them
 * and return the new list.
 */
nodeptr append(nodeptr a, nodeptr b)
{
    nodeptr aLast, bLast;

    if(a == NULL) return b;
    if(b == NULL) return a;
   
    aLast = a->left;
    bLast = b->left;

    join(aLast,b);
    join(bLast,a);

    return a;
}
/*
 * Convert given BST to linked list
 */
nodeptr treeToList(nodeptr root)
{
    nodeptr aList, bList;
    if(root == NULL)
        return NULL;
    else {
        aList = treeToList(root->left);
        bList = treeToList(root->right);
       
        root->left = root;
        root->right = root;
       
        aList = append(aList, root);
        aList = append(aList, bList);

        return aList;
    }
}
/*
 * print circular doubly linked list
 */
void printList(nodeptr head)
{
    nodeptr temp = head;
    if(head != NULL) {
        while(head->right != temp) {
            printf("%d ", head->data);
            head = head->right;
        }
        printf("%d ", head->data);
    }
    printf("\n");
}

Given a Binary Tree check whether it is Binary Search Tree or not

Write an isBST() function that returns true if a tree is a binary search tree
and false otherwise.

/*
 * Is BST helper function
 */
int isBSTUtil(nodeptr node, int min, int max)
{
    if(node != NULL)
        return (node->data >= min && node->data <=max) &&
            isBSTUtil(node->left, min, node->data) &&
            isBSTUtil(node->right, node->data, max);
    return 1;
           
}
/*
 * determine given binary tree is BST
 */
int isBST(nodeptr node)
{
    int min,max;
    if (node != NULL) {
        min = minValue(node);
        max = maxValue(node);
        return isBSTUtil(node, min, max);
    }
    return 1;
}

Check given two binary trees are same or not

Given two binary trees, return true if they are structurally identical -- they are made of nodes with the same values arranged in the same way.



/*
 * Compare two binary tree and return true if they are
 * structurally identical.
 */
int sameTree(nodeptr a, nodeptr b)
{
    if(a == NULL && b == NULL)
        return 1;
    else if((a == NULL && b != NULL) || (a != NULL && b == NULL))
        return 0;
    else {
        return (a->data == b->data && sameTree(a->left, b->left) &&
                sameTree(a->right, b->right));
    }
}

mirror()

Change a tree so that the roles of the left and right pointers are swapped at every node.




/*
 * Mirror image of given tree
 */
void mirror(nodeptr node)
{
    nodeptr temp;
    if(node != NULL)
    {
        temp = node->left;
        node->left = node->right;
        node->right = temp;
        mirror(node->left);
        mirror(node->right);
    }
}

Print all paths of Binary Tree

Given a binary tree, print out all of its root-to-leaf paths.

/*
 * Print array
 */
void printArray(int *path, int pathlen)
{
    int i;
    for(i = 0; i < pathlen; i++) {
        printf("%d ",path[i]);
    }
    printf("\n");
}
/*
 * Helper function to print all paths
 */
void printPathsRecur(nodeptr node, int path[], int pathLen)
{
    path[pathLen] = node->data;
    pathLen++;
    if(node->left == NULL && node->right == NULL) {
        printArray(path, pathLen);
    } else {
        printPathsRecur(node->left, path, pathLen);
        printPathsRecur(node->right, path, pathLen);
    }
}
/*
 * Print all paths
 */
void printPaths(nodeptr node)
{
    int path[200];
    if(node != NULL)
    {
        printPathsRecur(node, path, 0);
    }
}

hasPathSum - Binary Tree

Given a binary tree and a sum, return true if the tree has a root-to-leaf path such that adding up all the values
along the path equals the given sum. Return false if no such path can be found.

So for example, the following tree has exactly four root-to-leaf paths:
                       5
                   /        \
                4           8
              /          /       \         
          11         13        4
         /     \                     \
       7       2                    1

Root-to-leaf paths:
path 1: 5 4 11 7
path 2: 5 4 11 2
path 3: 5 8 13
path 4: 5 8 4 1
For this problem, we will be concerned with the sum of the values of such a path -- for example, the sum of the values on the 5-4-11-7 path is 5 + 4 + 11 + 7 = 27.

/*
 * "root-to-leaf path" to be a sequence of nodes in a tree
 * starting with the root node and proceeding downward to a leaf.
 */
int hasPathSum(nodeptr node, int sum)
{
    int subsum;
    if(node == NULL) {
        return (sum == 0);
    } else {
        subsum = sum - node->data;
        return (hasPathSum(node->left, subsum) ||
                hasPathSum(node->right, subsum));
    }
}