I am looking for a function which counts how many numbers in a range of cells are in the set of numbers
For example I have the set of numbers(1,2,3) and my cells contains 1 | 2 | 3 | 4 | 5 | 3 , the count should return 4
I have tried using countif but no success, I would like to have an excel function Ex.: =countif(A1:D5,...)
How about this? Assume data is in range A1:D5 and you want to count cells with a value of 1, 2 or 3:
=SUM(COUNTIF(A1:D5, {"1","2","3"}))
I hope my pseudo-code would be understandable
int count(int *set, int set_size, int *cells, int cells_size)
{
int v = 0;
// For every number in set
for(int i = 0; i < set_size; ++i)
{
// Loop through every number in cells
for(int j = 0; j < cells_size; ++j)
{
// If number in cells equals number in set, increment v
if(cells[j] == set[i])
{
v++;
}
}
}
// Result is in v, return it
return v;
}
Of course you can optimize a bit with using better containers than just arrays and sizes of them, but I hope you get the basics from this.
Note I used C-like language for pseudo-code, if anything is unclear I can explain further.
Related
I'm facing difficulty in understanding O(sum) complexity solution of coin changing problem.
The problem statement is:
You are given a set of coins A. In how many ways can you make sum B assuming you have infinite amount of each coin in the set.
NOTE:
Coins in set A will be unique. Expected space complexity of this problem is O(B).
The solution is:
int count( int S[], int m, int n )
{
int table[n+1];
memset(table, 0, sizeof(table));
table[0] = 1;
for(int i=0; i<m; i++)
for(int j=S[i]; j<=n; j++)
table[j] += table[j-S[i]];
return table[n];
}
can someone explain me this code.?
First, let's identify the parameters and variables used in the function:
Parameters:
S contain the denomination of all m coins. i.e. Each element contain the value of each coin.
m represents the number of coin denominations. Essentially, it's the length of array S.
n represents the sum B to be achieved.
Variables:
table: Element i in array table contains the number of ways sum i can be achieved with the given coins. table[0] = 1 because there is a single way to achieve a sum of 0 (not using any coin).
i loops through each coin.
Logic:
The number of ways to achieve a sum j = sum of the following:
number of ways to achieve a sum of j - S[0]
number of ways to achieve a sum of j - S[1]
...
number of ways to achieve a sum of j - S[m-1] (S[m-1] is the value of the mth coin)
I did not completely decipher nor validate the rest of the code, but I hope this is a step in the right direction.
Added comments to code:
#include <stdio.h>
#include <string.h>
int count( int S[], int m, int n )
{
int table[n+1];
memset(table, 0, sizeof(table));
table[0] = 1;
for(int i=0; i<m; i++) // Loop through all of the coins
for(int j=S[i]; j<=n; j++) // Achieve sum j between the value of S[i] and n.
table[j] += table[j-S[i]]; // Add to the number of ways to achieve sum j the number of ways to achieve sum j - S[i]
return table[n];
}
int main() {
int S[] = {1, 2};
int m = 2;
int n = 3;
int c = count(S, m, n);
printf("%d\n", c);
}
Notes:
The code avoids repeats: 3 = 1+1+1, 1+2 (2 ways instead of 3 if 2+1 was considered.
No dependence on the order of the coins in term of value.
We need to find the maximum element in an array which is also equal to product of two elements in the same array. For example [2,3,6,8] , here 6=2*3 so answer is 6.
My approach was to sort the array and followed by a two pointer method which checked whether the product exist for each element. This is o(nlog(n)) + O(n^2) = O(n^2) approach. Is there a faster way to this ?
There is a slight better solution with O(n * sqrt(n)) if you are allowed to use O(M) memory M = max number in A[i]
Use an array of size M to mark every number while you traverse them from smaller to bigger number.
For each number try all its factors and see if those were already present in the array map.
Here is a pseudo code for that:
#define M 1000000
int array_map[M+2];
int ans = -1;
sort(A,A+n);
for(i=0;i<n;i++) {
for(j=1;j<=sqrt(A[i]);j++) {
int num1 = j;
if(A[i]%num1==0) {
int num2 = A[i]/num1;
if(array_map[num1] && array_map[num2]) {
if(num1==num2) {
if(array_map[num1]>=2) ans = A[i];
} else {
ans = A[i];
}
}
}
}
array_map[A[i]]++;
}
There is an ever better approach if you know how to find all possible factors in log(M) this just becomes O(n*logM). You have to use sieve and backtracking for that
#JerryGoyal 's solution is correct. However, I think it can be optimized even further if instead of using B pointer, we use binary search to find the other factor of product if arr[c] is divisible by arr[a]. Here's the modification for his code:
for(c=n-1;(c>1)&& (max==-1);c--){ // loop through C
for(a=0;(a<c-1)&&(max==-1);a++){ // loop through A
if(arr[c]%arr[a]==0) // If arr[c] is divisible by arr[a]
{
if(binary_search(a+1, c-1, (arr[c]/arr[a]))) //#include<algorithm>
{
max = arr[c]; // if the other factor x of arr[c] is also in the array such that arr[c] = arr[a] * x
break;
}
}
}
}
I would have commented this on his solution, unfortunately I lack the reputation to do so.
Try this.
Written in c++
#include <vector>
#include <algorithm>
using namespace std;
int MaxElement(vector< int > Input)
{
sort(Input.begin(), Input.end());
int LargestElementOfInput = 0;
int i = 0;
while (i < Input.size() - 1)
{
if (LargestElementOfInput == Input[Input.size() - (i + 1)])
{
i++;
continue;
}
else
{
if (Input[i] != 0)
{
LargestElementOfInput = Input[Input.size() - (i + 1)];
int AllowedValue = LargestElementOfInput / Input[i];
int j = 0;
while (j < Input.size())
{
if (Input[j] > AllowedValue)
break;
else if (j == i)
{
j++;
continue;
}
else
{
int Product = Input[i] * Input[j++];
if (Product == LargestElementOfInput)
return Product;
}
}
}
i++;
}
}
return -1;
}
Once you have sorted the array, then you can use it to your advantage as below.
One improvement I can see - since you want to find the max element that meets the criteria,
Start from the right most element of the array. (8)
Divide that with the first element of the array. (8/2 = 4).
Now continue with the double pointer approach, till the element at second pointer is less than the value from the step 2 above or the match is found. (i.e., till second pointer value is < 4 or match is found).
If the match is found, then you got the max element.
Else, continue the loop with next highest element from the array. (6).
Efficient solution:
2 3 8 6
Sort the array
keep 3 pointers C, B and A.
Keeping C at the last and A at 0 index and B at 1st index.
traverse the array using pointers A and B till C and check if A*B=C exists or not.
If it exists then C is your answer.
Else, Move C a position back and traverse again keeping A at 0 and B at 1st index.
Keep repeating this till you get the sum or C reaches at 1st index.
Here's the complete solution:
int arr[] = new int[]{2, 3, 8, 6};
Arrays.sort(arr);
int n=arr.length;
int a,b,c,prod,max=-1;
for(c=n-1;(c>1)&& (max==-1);c--){ // loop through C
for(a=0;(a<c-1)&&(max==-1);a++){ // loop through A
for(b=a+1;b<c;b++){ // loop through B
prod=arr[a]*arr[b];
if(prod==arr[c]){
System.out.println("A: "+arr[a]+" B: "+arr[b]);
max=arr[c];
break;
}
if(prod>arr[c]){ // no need to go further
break;
}
}
}
}
System.out.println(max);
I came up with below solution where i am using one array list, and following one formula:
divisor(a or b) X quotient(b or a) = dividend(c)
Sort the array.
Put array into Collection Col.(ex. which has faster lookup, and maintains insertion order)
Have 2 pointer a,c.
keep c at last, and a at 0.
try to follow (divisor(a or b) X quotient(b or a) = dividend(c)).
Check if a is divisor of c, if yes then check for b in col.(a
If a is divisor and list has b, then c is the answer.
else increase a by 1, follow step 5, 6 till c-1.
if max not found then decrease c index, and follow the steps 4 and 5.
Check this C# solution:
-Loop through each element,
-loop and multiply each element with other elements,
-verify if the product exists in the array and is the max
private static int GetGreatest(int[] input)
{
int max = 0;
int p = 0; //product of pairs
//loop through the input array
for (int i = 0; i < input.Length; i++)
{
for (int j = i + 1; j < input.Length; j++)
{
p = input[i] * input[j];
if (p > max && Array.IndexOf(input, p) != -1)
{
max = p;
}
}
}
return max;
}
Time complexity O(n^2)
this is what i have of the function so far. This is only the beginning of the problem, it is asking to generate the random numbers in a 10 by 5 group of numbers for the output, then after this it is to be sorted by number size, but i am just trying to get this first part down.
/* Populate the array with 50 randomly generated integer values
* in the range 1-50. */
void populateArray(int ar[], const int n) {
int n;
for (int i = 1; i <= length - 1; i++){
for (int i = 1; i <= ARRAY_SIZE; i++) {
i = rand() % 10 + 1;
ar[n]++;
}
}
}
First of all we want to use std::array; It has some nice property, one of which is that it doesn't decay as a pointer. Another is that it knows its size. In this case we are going to use templates to make populateArray a generic enough algorithm.
template<std::size_t N>
void populateArray(std::array<int, N>& array) { ... }
Then, we would like to remove all "raw" for loops. std::generate_n in combination with some random generator seems a good option.
For the number generator we can use <random>. Specifically std::uniform_int_distribution. For that we need to get some generator up and running:
std::random_device device;
std::mt19937 generator(device());
std::uniform_int_distribution<> dist(1, N);
and use it in our std::generate_n algorithm:
std::generate_n(array.begin(), N, [&dist, &generator](){
return dist(generator);
});
Live demo
I'm Creating a game similar to Gift Quest, Where I have to swap elements, Check for Vertical and Horizontal repeated elements then erase repeated ones (if 3 elements are same in a row or column, remove those and fill with new ones)
int [][] Board = new int[5][5];
I have an integer array as my board i have defined each gift with a specific integer value for example chocolates = 1, candy = 2 etc
for(int i=0;i<5;i++)
{
for(int j=0;j<5;j++)
{
Board[i][j] = rand.nextInt(5);
}
}
I finished the swap part also, Now I want to check the board if there are same element more than 3 times in a row and column.if it is then make that place in array blank. So i can fill it.
Can someone help me with the checking part ?
int counter = 0;
int currentTile = 0;
for(int i=0;i<5;i++)
{
for(int j=0;j<5;j++)
{
if(Board[i,j] == currentTile)
Counter++;
if(Counter == 3)
DoStuff(); //Yay 3 in a row!
currentTile = Board[i,j];
}
counter = 0;
}
And do the same thing, but swap the loops for the x axis
How to autofit content in cell using jxl api?
I know this is an old question at this point, but I was looking for the solution to this and thought I would post it in case someone else needs it.
CellView Auto-Size
I'm not sure why the FAQ doesn't mention this, because it very clearly exists in the docs.
My code looked like the following:
for(int x=0;x<c;x++)
{
cell=sheet.getColumnView(x);
cell.setAutosize(true);
sheet.setColumnView(x, cell);
}
c stores the number of columns created
cell is just a temporary place holder for the returned CellView object
sheet is my WriteableSheet object
The Api warns that this is a processor intensive function, so it's probably not ideal for large files. But for a small file like mine (<100 rows) it took no noticeable time.
Hope this helps someone.
The method is self explanatory and commented:
private void sheetAutoFitColumns(WritableSheet sheet) {
for (int i = 0; i < sheet.getColumns(); i++) {
Cell[] cells = sheet.getColumn(i);
int longestStrLen = -1;
if (cells.length == 0)
continue;
/* Find the widest cell in the column. */
for (int j = 0; j < cells.length; j++) {
if ( cells[j].getContents().length() > longestStrLen ) {
String str = cells[j].getContents();
if (str == null || str.isEmpty())
continue;
longestStrLen = str.trim().length();
}
}
/* If not found, skip the column. */
if (longestStrLen == -1)
continue;
/* If wider than the max width, crop width */
if (longestStrLen > 255)
longestStrLen = 255;
CellView cv = sheet.getColumnView(i);
cv.setSize(longestStrLen * 256 + 100); /* Every character is 256 units wide, so scale it. */
sheet.setColumnView(i, cv);
}
}
for(int x=0;x<c;x++)
{
cell=sheet.getColumnView(x);
cell.setAutosize(true);
sheet.setColumnView(x, cell);
}
It is fine, instead of scanning all the columns. Pass the column as a parameter.
void display(column)
{
Cell = sheet.getColumnView(column);
cell.setAutosize(true);
sheet.setColumnView(column, cell);
}
So when you wiill be displaying your text you can set the particular length. Can be helpfull for huge excel files.
From the JExcelApi FAQ
How do I do the equivilent of Excel's "Format/Column/Auto Fit Selection"?
There is no API function to do this for you. You'll need to write code that scans the cells in each column, calculates the maximum length, and then calls setColumnView() accordingly. This will get you close to what Excel does but not exactly. Since most fonts have variable width characters, to get the exact same value, you would need to use FontMetrics to calculate the maximum width of each string in the column. No one has posted code on how to do this yet. Feel free to post code to the Yahoo! group or send it directly to the FAQ author's listed at the bottom of this page.
FontMetrics presumably refers to java.awt.FontMetrics. You should be able to work something out with the getLineMetrics(String, Graphics) method I would have though.
CellView's autosize method doesn't work for me all the time. My way of doing this is by programatically set the size(width) of the column based on the highest length of data in the column. Then perform some mathematical operations.
CellView cv = excelSheet.getColumnView(0);
cv.setSize((highest + ((highest/2) + (highest/4))) * 256);
where highest is an int that holds the longest length of data in the column.
setAutosize() method WILL NOT WORK if your cell has over 255 characters. This is related to the Excel 2003 max column width specification: http://office.microsoft.com/en-us/excel-help/excel-specifications-and-limits-HP005199291.aspx
You will need to write your own autosize method to handle this case.
Try this exemple:
expandColumns(sheet, 3);
workbook.write();
workbook.close();
private void expandColumn(WritableSheet sheet, int amountOfColumns){
int c = amountOfColumns;
for(int x=0;x<c;x++)
{
CellView cell = sheet.getColumnView(x);
cell.setAutosize(true);
sheet.setColumnView(x, cell);
}
}
Kotlin's implementation
private fun sheetAutoFitColumns(sheet: WritableSheet, columnsIndexesForFit: Array<Int>? = null, startFromRowWithIndex: Int = 0, excludeLastRows : Int = 0) {
for (columnIndex in columnsIndexesForFit?.iterator() ?: IntProgression.fromClosedRange(0, sheet.columns, 1).iterator()) {
val cells = sheet.getColumn(columnIndex)
var longestStrLen = -1
if (cells.isEmpty()) continue
for (j in startFromRowWithIndex until cells.size - excludeLastRows) {
if (cells[j].contents.length > longestStrLen) {
val str = cells[j].contents
if (str == null || str.isEmpty()) continue
longestStrLen = str.trim().length
}
}
if (longestStrLen == -1) continue
val newWidth = if (longestStrLen > 255) 255 else longestStrLen
sheet.setColumnView(columnIndex, newWidth)
}
}
example for use
sheetAutoFitColumns(sheet) // fit all columns by all rows
sheetAutoFitColumns(sheet, arrayOf(0, 3))// fit A and D columns by all rows
sheetAutoFitColumns(sheet, arrayOf(0, 3), 5)// fit A and D columns by rows after 5
sheetAutoFitColumns(sheet, arrayOf(0, 3), 5, 2)// fit A and D columns by rows after 5 and ignore two last rows