We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Digit Operations to Make Two Integers Equal

Number: 3655

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Given two integers n and m having the same number of digits, you may change any one digit of n at a time by either increasing it by 1 (if it is not 9) or decreasing it by 1 (if it is not 0). However, at every step (including the original n and at every intermediate result) the current integer must not be a prime number. The goal is to convert n to m with a sequence of these digit operations such that the transformation cost––which is defined as the sum of all the integer values that n takes (including the starting value and after each operation)––is minimized. Return the minimum cost if possible, otherwise return -1.

Key Insights

  • Model the problem as a weighted graph with nodes representing valid integers (with the given number of digits and not prime) and edges representing a single allowed digit operation.
  • The weight of an edge from one integer to its neighbor is the neighbor’s integer value (and the starting integer is also charged).
  • Use Dijkstra’s algorithm in order to find the minimum cost path from n to m.
  • Pre-compute all primes up to 10^4 (the maximum possible value) so that each generated neighbor can be quickly validated.
  • Ensure that digit operations do not produce numbers with a different number of digits (avoid producing a leading zero).

Space and Time Complexity

Time Complexity: O(N * D) where N is the number of states (roughly up to 9 * 10^(d-1) for numbers with d digits, d ≤ 4) and D is the number of digits. This is acceptable given the constraints. Space Complexity: O(N) for storing the cost for each valid state and the prime-check table.

Solution

We use Dijkstra’s algorithm to traverse the state space where each state is a valid integer with the same number of digits as n. Each transition changes one digit by ±1 (ensuring that the most significant digit never becomes 0 and that intermediate states are not prime). The cost accumulated when moving to a neighbor equals the neighbor’s value. The algorithm proceeds by:

  1. Precomputing a sieve to mark prime numbers up to 10^4.
  2. Checking that the starting integer (n) and target integer (m) are non-prime.
  3. Converting the current integer to a string representation so that we can perform digit-level operations.
  4. For each valid digit operation (increase if digit < 9, decrease if digit > 0 and if it doesn’t result in a leading zero), computing the new integer value and validating that it is non-prime.
  5. Using a min-heap (priority queue) to always expand the state with the lowest total cost path so far.
  6. If m is reached, return the total cost; otherwise, if no valid transformation exists, return -1.

Code Solutions

import heapq

def is_prime_sieve(limit):
    sieve = [True] * (limit+1)
    sieve[0] = sieve[1] = False
    for i in range(2, int(limit**0.5)+1):
        if sieve[i]:
            for j in range(i*i, limit+1, i):
                sieve[j] = False
    return sieve

def minCostTransformation(n, m):
    # Determine number of digits and bounds
    str_n = str(n)
    digits = len(str_n)
    low_bound = 10**(digits-1)
    high_bound = 10**digits - 1
    
    # Pre-compute prime information for numbers up to high_bound (or 10^4)
    sieve = is_prime_sieve(10000)
    
    # If n or m is prime, transformation is invalid
    if sieve[n] or sieve[m]:
        return -1
    
    # Dijkstra: each state is (cost, value)
    heap = [(n, n)]  # (current total cost, current value)
    # Use dictionary to store the minimum cost to reach state.
    best = {n: n}
    
    while heap:
        cost, cur = heapq.heappop(heap)
        # Check if we reached target state m
        if cur == m:
            return cost
        # If we already found a cheaper path, skip processing this state.
        if cost > best.get(cur, float('inf')):
            continue
        cur_str = str(cur)
        # Explore all neighbors by adjusting each digit by ±1
        for i in range(digits):
            digit = int(cur_str[i])
            for d in (-1, 1):
                new_digit = digit + d
                # Check digit constraints: within [0,9]
                if new_digit < 0 or new_digit > 9:
                    continue
                # For the most significant digit, ensure it is not 0
                if i == 0 and new_digit == 0:
                    continue
                # Create new number string with one digit modified
                new_num_str = cur_str[:i] + str(new_digit) + cur_str[i+1:]
                new_num = int(new_num_str)
                # Must have same number of digits and be non-prime
                if new_num < low_bound or new_num > high_bound or sieve[new_num]:
                    continue
                new_cost = cost + new_num
                if new_cost < best.get(new_num, float('inf')):
                    best[new_num] = new_cost
                    heapq.heappush(heap, (new_cost, new_num))
    return -1

# Example usage:
print(minCostTransformation(10, 12))  # Expected output: 85
print(minCostTransformation(4,8))     # Expected output: -1
print(minCostTransformation(6,2))     # Expected output: -1
← Back to All Questions