Sample case:
a[]={1,4,2,5,10,5,4}; //output== 9
here we cannot change the order of element smaller always appear before bigger element
The constraint are:
1<=n<=10^4
10^-6<=a[i]<=10^-6
Here is my code but it will fail on some test cases ,can someone find the error in this code.
Thank You .
int maxDiff(int *a,int n)
{
int MIN_value=INT_MAX;
int MAX_vale=INT_MIN;
for(int i=0;i<n;i++) {
MIN_value=min(MIN_value,a[i]);
MAX_value=max(a[i]-MIN_value,MAX_value);
}
return MAX_value;
}
Your function is wrong because, if it finds the minimum value at the end of the array, all your previous calculations of differences are invalidated. Your error is thus in the line:
MAX_value=max(a[i]-MIN_value,MAX_value);
A much better way to go about this would be:
int maxDiff(int *a,int n)
{
if (n == 0 || n == 1) //If list is empty or has 1 element, return 0
return 0;
int MIN_value=INT_MAX;
int MAX_vale=INT_MIN;
for(int i=0;i<n;i++) {
MIN_value=min(MIN_value,a[i]);
MAX_value=max(a[i],MAX_value);
}
return MAX_value - MIN_VALUE;
}
The error must be in the logic of this line:
MAX_value=max(a[i]-MIN_value,MAX_value);
Use a debugger to test the method on different inputs and trace if MAX_Value gets the values assigned you expect it to get.
You may just set MAX_value correctly and return the difference afterwards:
MAX_value=max(MAX_value,a[i]);
Finally:
return MAX_VALUE - MIN_VALUE;
Be aware that if the difference is negative then the array was empty.
Edit: Compute the difference between the maximal value and the minimal value to the left of the maximal value (see comments):
possibleMinValue = min(possibleMinValue,a[i]);
oldMaxValue = MAX_value;
MAX_value=max(MAX_value,a[i]);
if (oldMaxValue != MAX_value) {
// Found a new maximal value. Thus, possibleMinValue is a valid min value
MIN_value = possibleMinValue;
}
When finding the min or max of an array it is best to set the initial max or min to the first possible outcome; in this case it would be number 1 - number 2. This is to ensure that INT_MAX and INT_MIN do not become your final answer. Alternatively, when finding the min you can use INT_MAX as the initial value for your min or vice versa.
int findMaxDiff(int* array, int n) {
int max = array[i];
int min = array[i];
for(int i = 0; i < n; i++) {
if(array[i] > max) {
max = array[i]
}
if(array[i] < min) {
min = array[i]
}
}
return max - min;
}
Related
I have got an assignment and i'll be glad if you can help me with one question
in this assignment, i have a question that goes like this:
write a function that receives an array and it's length.
the purpose of the function is to check if the array has all numbers from 0 to length-1, if it does the function will return 1 or 0 otherwise.The function can go through the array only one.
you cant sort the array or use a counting array in the function
i wrote the function that calculate the sum and the product of the array's values and indexes
int All_Num_Check(int *arr, int n)
{
int i, index_sum = 0, arr_sum = 0, index_multi = 1, arr_multi = 1;
for (i = 0; i < n; i++)
{
if (i != 0)
index_multi *= i;
if (arr[i] != 0)
arr_multi *= arr[i];
index_sum += i;
arr_sum += arr[i];
}
if ((index_sum == arr_sum) && (index_multi == arr_multi))
return 1;
return 0;
}
i.e: length = 5, arr={0,3,4,2,1} - that's a proper array
length = 5 , arr={0,3,3,4,2} - that's not proper array
unfortunately, this function doesnt work properly in all different cases of number variations.
i.e: length = 5 , {1,2,2,2,3}
thank you your help.
Checking the sum and product is not enough, as your counter-example demonstrates.
A simple solution would be to just sort the array and then check that at every position i, a[i] == i.
Edit: The original question was edited such that sorting is also prohibited. Assuming all the numbers are positive, the following solution "marks" numbers in the required range by negating the corresponding index.
If any array cell already contains a marked number, it means we have a duplicate.
int All_Num_Check(int *arr, int n) {
int i, j;
for (i = 0; i < n; i++) {
j = abs(arr[i]);
if ((j >= n) || (arr[j] < 0)) return 0;
arr[j] = -arr[j];
}
return 1;
}
I thought for a while, and then i realized that it is a highly contrained problem.
Things that are not allowed:
Use of counting array.
Use of sorting.
Use of more than one pass to the original array.
Hence, i came up with this approach of using XOR operation to determine the results.
a ^ a = 0
a^b^c = a^c^b.
Try this:
int main(int argc, char const *argv[])
{
int arr[5], i, n , temp = 0;
for(i=0;i<n; i++){
if( i == 0){
temp = arr[i]^i;
}
else{
temp = temp^(i^arr[i]);
}
}
if(temp == 0){
return 1;
}
else{
return 0;
}
}
To satisfy the condition mentioned in the problem, every number has to occour excatly once.
Now, as the number lies in the range [0,.. n-1], the looping variable will also have the same possible range.
Variable temp , is originally set to 0.
Now, if all the numbers appear in this way, then each number will appear excatly twice.
And XORing the same number twice results in 0.
So, if in the end, when the whole array is traversed and a zero is obtained, this means that the array contains all the numbers excatly once.
Otherwise, multiple copies of a number is present, hence, this won't evaluate to 0.
Hey guys started programming in C few weeks ago learning about algothiritms, just wondering how would you make my code more simple its just a binary search function. But the only thing is you must keep the arguments the same, thanks in advance.
bool search(int value, int values[], int n)
{
int min = values[0];
int max = values[n-1];
int average = (min + max) / 2;
if(average == value)
{
return true;
}
while (average > value)
{
max = average - 1;
average = (min + max) / 2;
}
while (average < value)
{
min = average + 1;
average = (min + max) / 2;
}
if (max < min)
{
return false;
}
if (average == value) {
printf("%i\n", average);
return true;
}
else
{
return false;
}
}
There are a bunch of little things you have to get right in a binary search: handle the length=0 case, make sure the position you test is always valid, make sure you don't overflow (i.e., `(low+high)/2' is not the best way to write that), make sure the new test position is always different from the previous one, etc.
After having done it like a million times, every binary search I write is now done just like this:
bool search(int[] array, int length, int valueToFind)
{
int pos=0;
int limit=length;
while(pos<limit)
{
int testpos = pos+((limit-pos)>>1);
if (array[testpos]<valueToFind)
pos=testpos+1;
else
limit=testpos;
}
return (pos < length && array[pos]==valueToFind);
}
Notice that we only need to do one comparison per iteration, which is faster than most implementations that can do 2. Instead of doing the equality test inside the loop, we reliably find the position where the element to find belongs, using only one comparison per iteration, and then at the end test to see if the element we want is there.
The way we calculate testpos ensures that pos <= testpos < limit, AND it works even if length is the largest possible integer value.
This form also makes it very easy to read off the invariants you want to see, without having to think about strange boundary conditions like high<low. When you come out of the loop, pos==limit so you don't have to worry about using the wrong one, etc.
The condition in this loop is also easily adaptable to different-purpose binary searches like "find where to insert x, ensuring that it goes after all the xs that are already in the array", "find the first x in the array", "find the last x in the array", etc.
This is not a right implementation of binary search ,all conditions must be in one loop ,such as:
Also max must be n-1 and not values[n-1] and min=0 instead of values[0] as also you should compare values[average] with value not just average variable.
bool search(int value, int values[], int n){
int min = 0;
int max = n-1;
int average ;
while(max>=min){
average = (min + max) / 2;
if(values[average] == value)
{
return true;
}
if (values[average] > value)
{
max = average - 1;
average = (min + max) / 2;
}
if (values[average] < value)
{
min = average + 1;
average = (min + max) / 2;
}
}
return false;
}
I have a function that takes a one-dimensional array of N positive integers and returns the number of elements that are larger than all the next. The problem is exist a function to do it that in a better time? My code is the following:
int count(int *p, int n) {
int i, j;
int countNo = 0;
int flag = 0;
for(i = 0; i < n; i++) {
flag = 1;
for(j = i + 1; j < n; j++) {
if(p[i] <= p[j]) {
flag = 0;
break;
}
}
if(flag) {
countNo++;
}
}
return countNo;
}
My solution is O(n^2). Can it be done better?
You can solve this problem in linear time(O(n) time). Note that the last number in the array will always be a valid number that fits the problem definition. So the function will always output a value that will be greater than equal to 1.
For any other number in the array to be a valid number it must be greater than or equal to the greatest number that is after that number in the array.
So iterate over the array from right to left keeping track of the greatest number found till now and increment the counter if current number is greater than or equal to the greatest found till now.
Working code
int count2(int *p, int n) {
int max = -1000; //this variable represents negative infinity.
int cnt = 0;
int i;
for(i = n-1; i >=0; i--) {
if(p[i] >= max){
cnt++;
}
if(p[i] > max){
max = p[i];
}
}
return cnt;
}
Time complexity : O(n)
Space complexity : O(1)
It can be done in O(n).
int count(int *p, int n) {
int i, currentMax;
int countNo = 0;
currentMax = p[n-1];
for(i = n-1; i >= 0; i--) {
if(currentMax < p[i])
{
countNo ++;
currentMax = p[i];
}
}
return countNo;
}
Create an auxillary array aux:
aux[i] = max{arr[i+1], ... ,arr[n-1] }
It can be done in linear time by scanning the array from right to left.
Now, you only need the number of elements such that arr[i] > aux[i]
This is done in O(n).
Walk backwards trough the array, and keep track of the current maximum. Whenever you find a new maximum, that element is larger than the elements following.
Yes, it can be done in O(N) time. I'll give you an approach on how to go about it. If I understand your question correctly, you want the number of elements that are larger than all the elements that come next in the array provided the order is maintained.
So:
Let len = length of array x
{...,x[i],x[i+1]...x[len-1]}
We want the count of all elements x[i] such that x[i]> x[i+1]
and so on till x[len-1]
Start traversing the array from the end i.e. at i = len -1 and keep track of the largest element that you've encountered.
It could be something like this:
max = x[len-1] //A sentinel max
//Start a loop from i = len-1 to i = 0;
if(x[i] > max)
max = x[i] //Update max as you encounter elements
//Now consider a situation when we are in the middle of the array at some i = j
{...,x[j],....x[len-1]}
//Right now we have a value of max which is the largest of elements from i=j+1 to len-1
So when you encounter an x[j] that is larger than max, you've essentially found an element that's larger than all the elements next. You could just have a counter and increment it when that happens.
Pseudocode to show the flow of algorithm:
counter = 0
i = length of array x - 1
max = x[i]
i = i-1
while(i>=0){
if(x[i] > max){
max = x[i] //update max
counter++ //update counter
}
i--
}
So ultimately counter will have the number of elements you require.
Hope I was able to explain you how to go about this. Coding this should be a fun exercise as a starting point.
I have been attempting to code for a program that stores input into an array and then allows me to print it out. It also lets me know which number is the largest. What I am trying to figure out is how can I get my program to tell me the amount of times (occurrences) the largest number in array is input. Here is my code so far. As of now, this code outputs the numbers I enter to the array, the largest element in the array, and the occurrence of every number I input( The occurrences of the numbers are incorrect). In all the the amount of occurrences for every number turns out to be 0. Which is obviously incorrect. Again, I need my program to display the largest number (which it does) and the occurrences of ONLY the largest number. All advice, tips, or thoughts are welcome. Thank you.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
int main()
{
int arrayNum[15];
int a;
int max=0;
int location;
for( a=0; a < 15; a++)
{
printf("Enter element %d:", a);
scanf("%d",&arrayNum[a]);
}
for(a=0; a < 15; a++)
{
printf("%d\n", arrayNum[a]);
}
for (a = 1; a < 15; a++)
{
if (arrayNum[a] > max)
{
max = arrayNum[a];
location = a+1;
}
}
printf("Max element in the array in the location %d and its value %d\n", location, max);
for(a=0; a<15; a++)
{
if(arrayNum[a+1] == arrayNum[a])
continue;
else
printf("Number %d: %d occurences\n", arrayNum[a]);
}
return 0;
}
I spot some problems in your code. First, the third for loop starts at 1, but it does not update the max as the value of arrayNum[0].
Then, for the problem at hand, I would have two variables:
int max; // The maximum value
int max_count; // The count of the maximum value
Then, the logic to find the greatest, and the count, is the following:
For each element, compare it with the maximum seen. If it is equal, increment max_count. If it is bigger, update max with the value, and set the max_count to 1. If it is smaller, ignore it. Something like:
max = arrayNum[0];
max_count = 1;
for (int a = 1; a < 15; ++a)
{
if (arrayNum[a] == max)
max_count++;
else if (arrayNum[a] > max)
{
max_count = 1;
max = arrayNum[a];
}
}
All you need to do is introduce a new variable to keep track of the number of occurrences of max. When a new value of max is found, set that count to zero. When a subsequent value is found equal to the max, increment the counter.
Incidentally, your code doesn't properly find the maximum in its current form. Try one test case where your array elements are all negative. Try another test case in which all the values are positive, and the first value entered (arrayNum[0]) is the maximum. You will find, in both cases, that your function will not actually find the maximum.
Just before you begin the below loop max is still 0 make
max = a[0];
for (a = 1; a < 15; a++)
{
if (arrayNum[a] > max)
{
max = arrayNum[a];
location = a+1;
}
}
Later
int n=0;
for(i=0;i<15;i++)
{
if(max == a[i])
n++;
}
printf("Number of times max appears in the array is %d\n",n);
Replace last for loop with below code
NoOfOccurances = 0;
for(a=0; a<15; a++)
{
if(max == arrayNum[a])
{
NoOfOccurances++;
}
}
printf("Number %d: %d occurences\n", max,NoOfOccurances);
For your third for-loop, the one where you find out the largest number in your array, I would suggest to set max to arrayNum[0], that way it will work even with negative numbers.
Then, to know how many occurrence of the highest number there is, you need a count variable that you increment (count++) each time a number of the array is equal to max. To do that you need another for-loop.
Good luck.
You can do what you want in just one loop iteration:
int count = 1;
int position = 0;
int max = arrayNum[0];
int N = 15;
int p;
for (p = 1; p < N; ++p)
{
if (arrayNum[p] > max) // Find a bigger number
{
max = arrayNum[p];
pos = p;
count = 1;
}
else if ( arrayNum[p] == max) // Another occurrences of the same number
count++;
}
A simple solution with time complexity of O(n)
int maxoccurence(int a[],int ar_size)
{
int max=a[0],count=0,i;
for(i=0;i<ar_size;i++)
{
if(a[i]==max)//counting the occurrence of maximum element
count++;
if(a[i]>max)//finding maximum number
{
max=a[i];
count=1;
}
}
printf("Maximum element in the array is %d\n",max);
return count;
}
I'm having issues with a sort method I wrote. It is supposed to find the max value, and replace the last value in the array with the max (and move that value to where the last value was).
I've ran gdb, and it looks like the if statement always executes, and for some reason max = values[0] always sets max to 0. Granted I am very new to C so I might be wrong about what is going on.
/**
* Sorts array of n values.
*/
void sort(int values[], int n)
{
// TODO: implement an O(n^2) sorting algorithm
int max; //hold the max value through the iteration
int replaced; //to hold the value at the end of the array
int replacedhash; //to hold the location of the max value
do
{
replaced = values[n];
max = values[0]; //reset max to 0 for new iteration
for(int i = 0; i<n ; i++)
{
//check if the next value is larger,
//then update max and replacedhash if it is
if (max < values[i])
{
max = values[i];
replacedhash = i;
}
}
values[replacedhash] = replaced; //next three lines swap the values
n--;
values[n] = max;
} while (n!=0);
}
And I would use this by running:
int main() {
int test[] = {3,5,2,5,6,100,4,46};
sort(test, 8);
printarray(test, 8);
}
Error 1: replaced = values[n-1];
Your example in the problem statement is:
int test[] = {3,5,2,5,6,100,4,46};
sort(test, 8);
So you'll then look at test[8], which is undefined behavior
Error 2: replacedhash
replacedhash will be uninitialized if the first element of the array is the max. And it will probably have an incorrect value on later loops when the first element is the max.
My thoughts:
It appears to me that you've overcomplicating the code. You probably should just find the index in the array that has the maximum value, and then do the swap. It'll be simpler.
void sort(int values[], int n) {
do {
// Find index of maximum value
int max = 0;
for(int i=0; i<n; i++)
if (values[max] < values[i])
max = i;
// Swap
int temp = values[max];
values[max] = values[n-1];
values[n-1] = temp;
n--;
} while (n != 0);
}