All sorts of data can be thought of as points in high dimensional space. For exa
ID: 672620 • Letter: A
Question
All sorts of data can be thought of as points in high dimensional space. For example, a web page might be thought of as a point where each component is the number of words about a particular topic. For a bank, the components of the point might be related to your credit score and your debt to cash ratio, etc.
No matter what a point represents, it's important to be able to compute how similar two points are. For this assignment:
Create arrays for storing two points, both with 10 dimensions
Ask the user to input the components of the two points
print the distance between the two points
Distance is computed the same as it is for points in 2 or 3 d:
Your program should output should look something like this:
Explanation / Answer
Maximum difference between two elements such that larger element appears after the smaller number
Given an array arr[] of integers, find out the difference between any two elements such that larger element appears after the smaller number in arr[].
Examples: If array is [2, 3, 10, 6, 4, 8, 1] then returned value should be 8 (Diff between 10 and 2). If array is [ 7, 9, 5, 6, 3, 2 ] then returned value should be 2 (Diff between 7 and 9)
Method 1 (Simple)
Use two loops. In the outer loop, pick elements one by one and in the inner loop calculate the difference of the picked element with every other element in the array and compare the difference with the maximum difference calculated so far.
#include<stdio.h>
/* The function assumes that there are at least two
elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff(int arr[], int arr_size)
{
int max_diff = arr[1] - arr[0];
int i, j;
for(i = 0; i < arr_size; i++)
{
for(j = i+1; j < arr_size; j++)
{
if(arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
}
/* Driver program to test above function */
int main()
{
int arr[] = {1, 2, 90, 10, 110};
printf("Maximum difference is %d", maxDiff(arr, 5));
getchar();
return 0;
}
Run on IDE
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Method 2 (Tricky and Efficient)
In this method, instead of taking difference of the picked element with every other element, we take the difference with the minimum element found so far. So we need to keep track of 2 things:
1) Maximum difference found so far (max_diff).
2) Minimum number visited so far (min_element).
#include<stdio.h>
/* The function assumes that there are at least two
elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff(int arr[], int arr_size)
{
int max_diff = arr[1] - arr[0];
int min_element = arr[0];
int i;
for(i = 1; i < arr_size; i++)
{
if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
}
return max_diff;
}
/* Driver program to test above function */
int main()
{
int arr[] = {1, 2, 6, 80, 100};
int size = sizeof(arr)/sizeof(arr[0]);
printf("Maximum difference is %d", maxDiff(arr, size));
getchar();
return 0;
}
Run on IDE
Time Complexity: O(n)
Auxiliary Space: O(1)
Like min element, we can also keep track of max element from right side. See below code suggested by Katamaran
int maxDiff(int arr[], int n)
{
int maxDiff = -1; // Initialize Result
int maxRight = arr[n-1]; // Initialize max element from right side
for (int i = n-2; i >= 0; i--)
{
if (arr[i] > maxRight)
maxRight = arr[i];
else
{
int diff = maxRight - arr[i];
if (diff > maxDiff)
{
maxDiff = diff;
}
}
}
return maxDiff;
}
Run on IDE
Method 3 (Another Tricky Solution)
First find the difference between the adjacent elements of the array and store all differences in an auxiliary array diff[] of size n-1. Now this problems turns into finding the maximum sum subarray of this difference array.
Thanks to Shubham Mittal for suggesting this solution.
#include<stdio.h>
int maxDiff(int arr[], int n)
{
// Create a diff array of size n-1. The array will hold
// the difference of adjacent elements
int diff[n-1];
for (int i=0; i < n-1; i++)
diff[i] = arr[i+1] - arr[i];
// Now find the maximum sum subarray in diff array
int max_diff = diff[0];
for (int i=1; i<n-1; i++)
{
if (diff[i-1] > 0)
diff[i] += diff[i-1];
if (max_diff < diff[i])
max_diff = diff[i];
}
return max_diff;
}
/* Driver program to test above function */
int main()
{
int arr[] = {80, 2, 6, 3, 100};
int size = sizeof(arr)/sizeof(arr[0]);
printf("Maximum difference is %d", maxDiff(arr, size));
return 0;
}
Run on IDE
Output:
This method is also O(n) time complexity solution, but it requires O(n) extra space
Time Complexity: O(n)
Auxiliary Space: O(n)
We can modify the above method to work in O(1) extra space. Instead of creating an auxiliary array, we can calculate diff and max sum in same loop. Following is the space optimized version.
int maxDiff (int arr[], int n)
{
// Initialize diff, current sum and max sum
int diff = arr[1]-arr[0];
int curr_sum = diff;
int max_sum = curr_sum;
for(int i=1; i<n-1; i++)
{
// Calculate current diff
diff = arr[i+1]-arr[i];
// Calculate current sum
if (curr_sum > 0)
curr_sum += diff;
else
curr_sum = diff;
// Update max sum, if needed
if (curr_sum > max_sum)
max_sum = curr_sum;
}
return max_sum;
}
Run on IDE
Time Complexity: O(n)
Auxiliary Space: O(1)
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem
#include<stdio.h>
/* The function assumes that there are at least two
elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff(int arr[], int arr_size)
{
int max_diff = arr[1] - arr[0];
int i, j;
for(i = 0; i < arr_size; i++)
{
for(j = i+1; j < arr_size; j++)
{
if(arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
}
/* Driver program to test above function */
int main()
{
int arr[] = {1, 2, 90, 10, 110};
printf("Maximum difference is %d", maxDiff(arr, 5));
getchar();
return 0;
}