print common elements between two arrays from jagged array and multidimensional array

Question

    int[][] arr1 = new int[][] {{1,2,1},{9,7,2},{7,3,6}};

    int arr2[][] = new int[][] {{2,6,8},{0,1,7},{7,2,0},{8,3}};

    boolean duplicate= false;

    for(int i=0;i<arr1.length;i++) { 
        for(int j=0;j<arr1[i].length;j++) {
            int transpose = 0;
            transpose = arr1[i][j];

            for(int k=0;k<arr2.length;k++) {
                duplicate = false; 
                for(int  l=0;l<arr2[k].length;l++) { 
                    if(transpose == arr2[k][l]) {
                        System.out.println(transpose);
                        duplicate = true;
                        break;
                    }
                }
            
                if(duplicate)
                    break;
            }
        }

    }

}

}

I tried to print common elements without using any shortcuts only logically with for loop and if statements. The logic is, in array 1 i am iterating each element and checking if it is there in array 2 if it is there we have to print it in console. But the problem is in array 1 element "1,2,7" is iterating 2 times and printing it 2 times.

getting this Output: 1 2 1 7 2 7 3 6

Expected Output: 1 2 7 3 6

Here, i have to stop iterating twice because it is already checked once printed. note: I want answer only with looping statements and conditional statements without hash set or shortcuts.

Answer 1

You'll need a loop for each array, and then a loop for each sub-array.

Additionally, I'm utilizing a Set here to prevent duplicate values.

Set<Integer> d = new TreeSet<>();
int i, ni, j, nj;
for (int[] x : arr1)
    for (i = 0, ni = x.length; i < ni; i++)
        for (int[] y : arr2)
            for (j = 0, nj = y.length; j < nj; j++)
                if (x[i] == y[j]) d.add(x[i]);

Output

[1, 2, 3, 6, 7]

Ideally though, just add each value to a List, checking the List, first.

List<Integer> d = new ArrayList<>();
int i, ni, j, nj;
for (int[] x : arr1)
    for (i = 0, ni = x.length; i < ni; i++)
        for (int[] y : arr2)
            for (j = 0, nj = y.length; j < nj; j++)
                if (x[i] == y[j] && !d.contains(x[i])) d.add(x[i]);

Answer 2

Note: This solution is assuming we must use only primitive types, loops, and if/else statements

import java.util.*;

public class Main {
  public static void main(String[] args) {
    // Note: If not limited to primitive types, loops, and if statements.
    //  A better option would be using a Set<int>. It would be a similar process
    //  to the below solution, but far more efficient as insertion and retrieval
    //  of values would be O(1) instead of O(n).

    // when limited to using only primitives, arrays, and if statements
    int[][] arr1 = new int[][] {{1,2,1},{9,7,2},{7,3,6}};
    int[][] arr2 = new int[][] {{2,6,8},{0,1,7},{7,2,0},{8,3}};

    // find one of the array's flattened size
    int seenSize = 0;
    for (int i = 0; i < arr1.length; i++) {
      for (int j = 0; j < arr1[i].length; j ++) {
        seenSize++;
      }
    }
    int[] seen = new int[seenSize];

    // iterate over the same array to create an array of any unique seen element
    int seenIndex = 0;
    for (int i = 0; i < arr1.length; i++) {
      for (int j = 0; j < arr1[i].length; j++) {
        if (!valueIsPresentIn(arr1[i][j], seen, seenIndex)) {
          seen[seenIndex] = arr1[i][j];
          seenIndex++;
        }
      }
    }

    // Note: sorting the list of seen items would make this next step more efficient, then binary search could be used.

    int[] printed = new int[seenSize];
    int printedIndex = 0;

    // iterate over the second array searching for a similar occurrence in the list of
    // seen values, if something is found and it has not been printed, we need to print
    // it and add to the list of already printed values
    for (int i = 0; i < arr2.length; i++) {
      for (int j = 0; j < arr2[i].length; j++) {
        if (valueIsPresentIn(arr2[i][j], seen, seenIndex)) {
          if (!valueIsPresentIn(arr2[i][j], printed, printedIndex)) {
            printed[printedIndex] = arr2[i][j];
            System.out.println(printed[printedIndex]);
            printedIndex++;
          }
        }
      }
    }
  }

  // Method to search for a value, this is unnecessary if no longer limited to
  // primitive types as there are built-in methods and countless libraries/solutions
  // to this.
  public static boolean valueIsPresentIn(int value, int[] array, int maxIndex) {
    for (int searchIndex = 0; searchIndex < maxIndex; searchIndex++) {
      if (array[searchIndex] == value) {
        return true;
      }
    }

    return false;
  }
}