Interview Question for Backend Developers


Team: 3
Country: India




Comment hidden because of low score. Click to expand.
0
of 0 vote

Type: Recursion + Loop Problem

Loop on String Length. That is, 1 to MAX(CEILING(N/K), (N-K)) (inclusive)
- Minimum String Length is 1 because non-empty string
- Maximum String Length is a bit tricky
-- K cuts would result in K+1 Strings
-- Case-A: Cuts are equal in size, so max str len is (N/K) (ceiled)
-- Case-B: Cuts are not equal. Max length of (K+1)th string is (N-K) where other K strings are of min len 1

Recursion on String

Psuedo-Code:

class StringCutter {
  private String str;
  private Stack<Pair<int, int>> substrings;
  private int numCuts;
  private int maxStringSize;
  private int maxCutSize;
  private int numPalindromes;
  
  public StringCutter(String input) {
    this.str = input;
    this.maxStringSize = input.Size();
    this.substring = new Stack<Pair<int, int>>();
  }
  
  public void cut(int cutCount) {
    this.substring.clear();
    this.numCuts = cutCount;
    this.maxCutSize = MAX(CEILING(maxStringSize/numCuts), (maxStringSize-numCuts));
    this.numPalindromes = 0;
    cutInternal(0);
    System.output.printf("Palindrome Count = %d", this.numPalindromes);
  }
  
  private void cutInternal(int usedOffset) {
    // Did we make enough cuts?
    if (substring.size() == this.numCuts) {
      
      // Do cut strings complete the string? if not discard
      if (usedOffset < maxStringSize) return;

      StringBuilder sb = new StringBuilder();
        
      for (Pair<int, int> p : substrings) {
        cutString = str.substring(p.key, p.value);
        if (isPalindrome(cutString)) this.numPalindromes++;
        sb.append(cutString + "|"));
      }
      
      System.output.println(sb.toString());
    }
  
    for (int len = 1; len <= maxCutSize; len++) {
      substring.push(new Pair<int,int>(usedOffset, len));
      cutInternal(usedOffset + len);
      substring.pop();
    }
  }
  
  private boolean isPalindrome(String s) {
   // Trivial code to check if 's' is a palindrome
  }
}

- Laxmi Narsimha Rao Oruganti November 26, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This is how I'd go about it,
K will denote the level of the string in the reverse order 0 will be the last
based on the level iterate from end of the previous level to N-level
the rest of the string from current end to the last will be recursively solved
if the current substring is palindrome send 1 (c+1) to the next level
the c will be added to count at the end will have the no of palindromes for the current split

static Scanner sc = new Scanner(System.in);
static String input;
static int count = 0;
static int n;

public static void main(String[] args) {
    System.out.print("Enter N and K : ");
    n = sc.nextInt();
    var k = sc.nextInt();
    input = sc.nextLine();
    System.out.print("Enter the String : ");
    input = sc.nextLine();
    checkPalin(0, 0, k);
    System.out.println("The count of Palindrome in " + input + " = " + count);
}

private static void checkPalin(int c, int start, int lvl) {
    if (lvl > 0)
        for (int end = start + 1; end <= n - lvl; end++) {
            int tmp = 0;
            if (isPalin(start, end)) tmp = 1;
            checkPalin(c + tmp, end, lvl - 1);
        }
    if (lvl == 0) {
        if (isPalin(start, n)) c += 1;
        count += c;
    }
}

private static boolean isPalin(int start, int end) {
    for (int i = 0; i < (end - start) / 2; i++)
        if (input.charAt(start + i) != input.charAt(end - i - 1))
            return false;
    return true;
}

- PeyarTheriyaa November 26, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

We could use dynamic programming. Something like this:

def count(s,k):  
  dp1 = [[0]*(len(s)+1) 
        for _ in range(k+1)] 
  dp2 = [[0]*(len(s)+1) 
        for _ in range(k+1)] 
  for i in range(len(s)):
    for j in range(i+1): 
      z = s[j:i+1]
      p = [0,1][z == z[::-1]]
      if j == 0:
        dp1[0][i+1] = p 
        dp2[0][i+1] = 1
      for c in range(1, min(k,j)+1): 
        w = dp1[c-1][j] 
        y = dp2[c-1][j]
        dp1[c][i+1] += w + p*y
        dp2[c][i+1] += y
  return dp1[k][len(s)]
print count ('aabbc', 2)

- adr November 27, 2018 | Flag Reply


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