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give code for this in java in the format shown in ...

Creado el: 12 de julio de 2025

Respondido usando O3 por Chat01

Pregunta

87819068_SKC1272025_18834.png87819068_SKC1272025_18834.png
87819068_SKC1272025_18857.png87819068_SKC1272025_18857.png

give code for this in java in the format shown in image

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img = Image.open('/mnt/data/87819068_SKC1272025_18834.png')
plt.imshow(img)
plt.axis('off')

(-0.5, 1365.5, 767.5, -0.5)

Python

python
img2 = Image.open('/mnt/data/87819068_SKC1272025_18857.png')
plt.imshow(img2)
plt.axis('off')

(-0.5, 1365.5, 767.5, -0.5)

Python

python
import numpy as np
plt.figure(figsize=(6,4))
plt.imshow(img2.crop((100,300,400,500)))

<matplotlib.image.AxesImage at 0x7ea1cec84e60>

Python

python
plt.imshow(img2.crop((100,450,400,650)))

<matplotlib.image.AxesImage at 0x7ea16decd9a0>

Python

python
plt.figure(figsize=(6,4))
plt.imshow(img2.crop((100,600,650,800)))

<matplotlib.image.AxesImage at 0x7ea16e10d220>

Python

python
plt.figure(figsize=(6,2))
plt.imshow(img2.crop((120,480,600,540)))
plt.axis('off')

(-0.5, 479.5, 59.5, -0.5)

Python

python
plt.figure(figsize=(6,2))
plt.imshow(img2.crop((100,510,600,580)))
plt.axis('off')

(-0.5, 499.5, 69.5, -0.5)

Python

python
plt.figure(figsize=(6,2))
plt.imshow(img2.crop((100,450,400,510)))
plt.axis('off')

(-0.5, 299.5, 59.5, -0.5)

Python

python
plt.figure(figsize=(6,2))
plt.imshow(img2.crop((100,475,400,530)))
plt.axis('off')

(-0.5, 299.5, 54.5, -0.5)

Pensamiento

Finalizing the multigram logic

Okay, after breaking down the string "xyyyxyyyx," I see it can be partitioned into substrings of size 3: "xyy", "yxy", and "yyx", which are all anagrams of each other, making 3 the valid substring length. The key idea is to find the smallest substring length (L) that can divide the original string into substrings that are anagrams of each other. If no such length exists, the answer is -1. I'll stick with this approach for the code.

Constructing the shortest anagram code

I’m thinking through the algorithm to solve the problem. First off, I’ll loop through substring lengths LL from 1 to n/2n/2. If the string length is divisible by LL, I’ll check if all substrings of that length have matching character frequencies. If they do, I’ll return the first substring. Otherwise, after checking all possibilities, I’ll return -1. The solution seems feasible with a complexity of O(N2)O(N^2), which should work for N<200N < 200. Now, I’ll write the code for this.

Thought for 1m 50s

Respuesta

Here’s the completed solution in the exact skeleton shown in the screenshots:

java
import java.util.*;

public class TestClass
{
    //  str is the input string
    public static String shortestAnagram(String str)
    {
        int n = str.length();

        // try every possible block length (must divide n and give ≥ 2 blocks)
        for (int len = 1; len <= n / 2; len++)
        {
            if (n % len != 0) continue;          // block size must divide n

            // frequency count for the first block
            int[] base = new int[26];
            for (int i = 0; i < len; i++)
                base[str.charAt(i) - 'a']++;

            boolean ok = true;

            // compare every subsequent block’s frequency with the first
            for (int start = len; start < n; start += len)
            {
                int[] cur = new int[26];
                for (int i = start; i < start + len; i++)
                    cur[str.charAt(i) - 'a']++;

                if (!Arrays.equals(base, cur))
                {
                    ok = false;
                    break;
                }
            }

            if (ok)                 // found the minimal-length block
                return str.substring(0, len);
        }
        return "-1";                // not a multigram
    }

    public static void main(String[] arg)
    {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine().trim();
        System.out.println(shortestAnagram(str));
        sc.close();
    }
}

How it works

  1. Iterate over every block length that divides the string (up to n/2, so at least two blocks will exist).
  2. Record character frequencies (base) for the first block.
  3. Slide through the rest of the string in blocks of that length, ensuring each has identical frequencies (Arrays.equals).
  4. The first length that satisfies the check is the shortest anagram; return its substring.
  5. If no length works, the word is not a multigram, so print -1.

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