Salesforce OA | SMTS

Hi Folks,

YOE: 6 Currently SSE @ Walmart

Recently gave OA for SMTS position at Salesforce India. Got the OA link after cold emailing a bunch of salesforce recruiters.

#1: Minimum Operations to Zero (Same as LeetCode problem #2571.) You are given an integer n. In one operation, you can either add or subtract any power of 2 (i.e., ±2⁰, ±2¹, ±2², ..., ±2ᵏ) to/from n.

Write a function that returns the minimum number of such operations required to reduce n to 0. public static int getMinOperations(int n)

Input: A single integer n (1 ≤ n ≤ 10⁹) Output: An integer representing the minimum number of operations to reduce n to 0.

Example: Input: 21 Output: 3

Explanation: 21 → 20 (subtract 1 = 2⁰)
20 → 16 (subtract 4 = 2²)
16 → 0 (subtract 16 = 2⁴)

#2: Integrity Score A data packet is represented as a string of lowercase English letters. The integrity score of the packet is defined as the number of substrings whose character values sum to a number divisible by the length of the substring.

Each character is assigned a value as follows: a,b -> 1 c,d,e -> 2 f,g,h -> 3 i,j,k -> 4 l,m,n -> 5 o,p,q -> 6 r,s,t -> 7 u,v,w -> 8 x,y,z -> 9

Write a function that returns the integrity score of the packet. public static int integrityScore(String dataPacket)

Input: A string dataPacket consisting of lowercase English letters (1 ≤ length ≤ 10⁵) Output: An integer representing the integrity score.

Example: Input: "cat" Output: 4

Solution:

public static int integrityScore(String dataPacket) {
    int n = dataPacket.length();
    int[] prefixSum = new int[n + 1];
    int count = 0;

    // Build prefix sum using custom mapping: 'a' = 1 etc
    for (int i = 0; i < n; i++) {
        char ch = dataPacket.charAt(i);
        prefixSum[i + 1] = prefixSum[i] + getValue(ch);
    }

    // Check all substrings
    for (int start = 0; start < n; start++) {
        for (int end = start + 1; end <= n; end++) {
            int length = end - start;
            int sum = prefixSum[end] - prefixSum[start];
            if (sum % length == 0) {
                count++;
            }
        }
    }

    return count;
}
  private static int getValue(char ch) {
    return switch (ch) {
      case 'a','b' -> 1;
      case 'c','d','e' -> 2;
      case 'f','g','h' -> 3;
      case 'i','j','k' -> 4;
      case 'l','m','n' -> 5;
      case 'o','p','q' -> 6;
      case 'r','s','t' -> 7;
      case 'u','v','w' -> 8;
      case 'x','y','z' -> 9;
      default -> 0;
    };
  }

Cleared and asked to appear for F2F rounds.