Mettl Coding Questions – 1

June 24, 2019

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)

Time Complexity: O(n^2)
Auxiliary Space: O(1)

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.

C

#include

/* 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;
}

JAVA

class MaximumDiffrence
{
/* 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 functions */
public static void main(String[] args)
{
MaximumDiffrence maxdif = new MaximumDiffrence();
int arr[] = {1, 2, 90, 10, 110};
System.out.println(“Maximum differnce is ” +
maxdif.maxDiff(arr, 5));
}

Tricky Solution but Efficient

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).

Time Complexity: O(n)
Auxiliary Space: O(1)

C

#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;

}

JAVA

class MaximumDiffrence
{
/* 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 functions */
public static void main(String[] args)
{
MaximumDiffrence maxdif = new MaximumDiffrence();
int arr[] = {1, 2, 90, 10, 110};
int size = arr.length;
System.out.println(“MaximumDifference is ” +
maxdif.maxDiff(arr, size));
}
}

 

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