Change algorithm of intPartitions method (recursive solution)

This commit is contained in:
2026-04-23 02:28:50 +03:30
parent 1389bf4749
commit 4a8f093faf
+28 -52
View File
@@ -1,4 +1,5 @@
import java.util.ArrayList; import java.util.ArrayList;
import java.util.List;
public class MainExercises public class MainExercises
{ {
@@ -150,64 +151,39 @@ public class MainExercises
public int[][] intPartitions(int n) { public int[][] intPartitions(int n) {
if(n == 1){ List<List<Integer>> allPartitions = new ArrayList<>();
return new int[][]{{1}}; ArrayList<Integer> current = new ArrayList<>();
} generatePartition(n, n, current, allPartitions);
ArrayList<int[]> result = new ArrayList<>(); // converting ArrayList of ArrayLists into a 2d array
int[] current = new int[n]; // current partition, to be built int[][] result = new int[allPartitions.size()][];
for (int i = 0; i < allPartitions.size(); i++) {
// we use stack arrays to have the situation saved. List<Integer> sublist = allPartitions.get(i);
int[] remainingStack = new int[n*n]; // remaining value, to be added result[i] = new int[sublist.size()];
// int[] maxValueStack = new int[n*n]; // maximum number we can use, to prevent repeating partitions for (int j = 0; j < sublist.size(); j++) {
int[] nextTryStack = new int[n*n]; // the next number we try result[i][j] = sublist.get(j);
int[] lenStack = new int[n*n]; // length of current partition, gives us the next index in current
int head = 0; // works like a pointer, it makes stack arrays make sense.
remainingStack[0] = n;
// maxValueStack[0] = n;
nextTryStack[0] = n;
lenStack[0] = 0;
while(head >= 0){
int remaining = remainingStack[head];
// int maxValue = maxValueStack[head];
int nextTry = nextTryStack[head];
int partitionLen = lenStack[head];
if(remaining == 0){ // partition has been completed
//
int[] partition = new int[partitionLen];
System.arraycopy(current, 0, partition, 0, partitionLen);
result.add(partition);
head--;
continue;
} }
if(nextTry <= 0){ // no choices left
head--;
continue;
}
int chosen = nextTry;
nextTryStack[head] = chosen - 1; // next value to be tried
current[partitionLen] = chosen; // adding selected number
// generating next number for the partition
head++;
remainingStack[head] = remaining - chosen;
// maxValueStack[head] = chosen;
nextTryStack[head] = Math.min(chosen, remaining - chosen);
lenStack[head] = partitionLen + 1;
} }
int[][] output = new int[result.size()][]; return result;
for(int i = 0; i < result.size(); i++){
output[i] = result.get(i); }
private void generatePartition(int remaining, int maxValue, List<Integer> currentPartition, List<List<Integer>> allPartitions){
if(remaining == 0){
allPartitions.add(new ArrayList<>(currentPartition));
return ;
} }
return output; for (int i = Math.min(maxValue, remaining); i >= 1; i--){ // preventing choosing the bigger number between maxValue and remaining, this way repeating partitions
currentPartition.add(i);
generatePartition(remaining - i, i, currentPartition, allPartitions);
currentPartition.removeLast(); /*
after the recursive call returns. we need to undo adding i.
this way we can try adding i again. ( remaining may not be 0 yet)
*/
}
} }