LeetCode #3463 — HARD

Check If Digits Are Equal in String After Operations II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s consisting of digits. Perform the following operation repeatedly until the string has exactly two digits:

For each pair of consecutive digits in s, starting from the first digit, calculate a new digit as the sum of the two digits modulo 10.
Replace s with the sequence of newly calculated digits, maintaining the order in which they are computed.

Return true if the final two digits in s are the same; otherwise, return false.

Example 1:

Input: s = "3902"

Output: true

Explanation:

Initially, s = "3902"
First operation:
- (s[0] + s[1]) % 10 = (3 + 9) % 10 = 2
- (s[1] + s[2]) % 10 = (9 + 0) % 10 = 9
- (s[2] + s[3]) % 10 = (0 + 2) % 10 = 2
- s becomes "292"
Second operation:
- (s[0] + s[1]) % 10 = (2 + 9) % 10 = 1
- (s[1] + s[2]) % 10 = (9 + 2) % 10 = 1
- s becomes "11"
Since the digits in "11" are the same, the output is true.

Example 2:

Input: s = "34789"

Output: false

Explanation:

Initially, s = "34789".
After the first operation, s = "7157".
After the second operation, s = "862".
After the third operation, s = "48".
Since '4' != '8', the output is false.

Constraints:

3 <= s.length <= 10⁵
s consists of only digits.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of digits. Perform the following operation repeatedly until the string has exactly two digits: For each pair of consecutive digits in s, starting from the first digit, calculate a new digit as the sum of the two digits modulo 10. Replace s with the sequence of newly calculated digits, maintaining the order in which they are computed. Return true if the final two digits in s are the same; otherwise, return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"3902"

Example 2

"34789"

Core Insight

What unlocks the optimal approach

Can we use <code>nCr</code> and use Pascal's triangle values here?
<code>nCr mod 10</code> can be uniquely determined from <code>nCr mod 2</code> and <code>nCr mod 5</code>.
Use Lucas's theorem.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
class Solution {
  // Same as 3461. Check If Digits Are Equal in String After Operations I
  public boolean hasSameDigits(String s) {
    final int n = s.length();
    int num1 = 0;
    int num2 = 0;

    for (int i = 0; i + 1 < n; ++i) {
      final int coefficient = nCMOD10(n - 2, i);
      num1 += (coefficient * (s.charAt(i) - '0')) % 10;
      num1 %= 10;
      num2 += (coefficient * (s.charAt(i + 1) - '0')) % 10;
      num2 %= 10;
    }

    return num1 == num2;
  }

  // Returns (n, k) % 10.
  private int nCMOD10(int n, int k) {
    final int mod2 = lucasTheorem(n, k, 2);
    final int mod5 = lucasTheorem(n, k, 5);
    int[][] lookup = {
        {0, 6, 2, 8, 4}, // mod2 == 0
        {5, 1, 7, 3, 9}  // mod2 == 1
    };
    return lookup[mod2][mod5];
  }

  // Returns (n, k) % prime.
  private int lucasTheorem(int n, int k, int prime) {
    int res = 1;
    while (n > 0 || k > 0) {
      final int nMod = n % prime;
      final int MOD = k % prime;
      res *= nCk(nMod, MOD);
      res %= prime;
      n /= prime;
      k /= prime;
    }
    return res;
  }

  // Returns (n, k).
  private int nCk(int n, int k) {
    int res = 1;
    for (int i = 0; i < k; ++i) {
      res *= (n - i);
      res /= (i + 1);
    }
    return res;
  }
}

// Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
package main

import "math/bits"

// https://space.bilibili.com/206214
const mod = 10

func pow(x, n int) int {
	res := 1
	for ; n > 0; n /= 2 {
		if n%2 > 0 {
			res = res * x % mod
		}
		x = x * x % mod
	}
	return res
}

const mx = 100_000

var f, invF, p2, p5 [mx + 1]int

func init() {
	f[0] = 1
	for i := 1; i <= mx; i++ {
		x := i
		// 2 的幂次
		e2 := bits.TrailingZeros(uint(x))
		x >>= e2
		// 5 的幂次
		e5 := 0
		for x%5 == 0 {
			e5++
			x /= 5
		}
		f[i] = f[i-1] * x % mod
		p2[i] = p2[i-1] + e2
		p5[i] = p5[i-1] + e5
	}

	invF[mx] = pow(f[mx], 3) // 欧拉定理
	for i := mx; i > 0; i-- {
		x := i
		x >>= bits.TrailingZeros(uint(x))
		for x%5 == 0 {
			x /= 5
		}
		invF[i-1] = invF[i] * x % mod
	}
}

var pow2 = [4]int{2, 4, 8, 6}

func comb(n, k int) int {
	res := f[n] * invF[k] * invF[n-k]
	e2 := p2[n] - p2[k] - p2[n-k]
	if e2 > 0 {
		res *= pow2[(e2-1)%4]
	}
	if p5[n]-p5[k]-p5[n-k] > 0 {
		res *= 5
	}
	return res
}

func hasSameDigits(s string) bool {
	diff := 0
	for i := range len(s) - 1 {
		diff += comb(len(s)-2, i) * (int(s[i]) - int(s[i+1]))
	}
	return diff%mod == 0
}

# Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
class Solution:
  # Same as 3461. Check If Digits Are Equal in String After Operations I
  def hasSameDigits(self, s: str) -> bool:
    n = len(s)
    num1 = 0
    num2 = 0

    for i in range(n - 1):
      coefficient = self._nCMOD10(n - 2, i)
      num1 += (coefficient * (int(s[i]) - 0)) % 10
      num1 %= 10
      num2 += (coefficient * (int(s[i + 1]) - 0)) % 10
      num2 %= 10

    return num1 == num2

  def _nCMOD10(self, n: int, k: int) -> int:
    """Returns (n, k) % 10."""
    mod2 = self._lucasTheorem(n, k, 2)
    mod5 = self._lucasTheorem(n, k, 5)
    lookup = [
        [0, 6, 2, 8, 4],  # mod2 == 0
        [5, 1, 7, 3, 9]   # mod2 == 1
    ]
    return lookup[mod2][mod5]

  def _lucasTheorem(self, n: int, k: int, prime: int) -> int:
    """Returns (n, k) % prime."""
    res = 1
    while n > 0 or k > 0:
      nMod = n % prime
      MOD = k % prime
      res *= math.comb(nMod, MOD)
      res %= prime
      n //= prime
      k //= prime
    return res

// Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
// class Solution {
//   // Same as 3461. Check If Digits Are Equal in String After Operations I
//   public boolean hasSameDigits(String s) {
//     final int n = s.length();
//     int num1 = 0;
//     int num2 = 0;
// 
//     for (int i = 0; i + 1 < n; ++i) {
//       final int coefficient = nCMOD10(n - 2, i);
//       num1 += (coefficient * (s.charAt(i) - '0')) % 10;
//       num1 %= 10;
//       num2 += (coefficient * (s.charAt(i + 1) - '0')) % 10;
//       num2 %= 10;
//     }
// 
//     return num1 == num2;
//   }
// 
//   // Returns (n, k) % 10.
//   private int nCMOD10(int n, int k) {
//     final int mod2 = lucasTheorem(n, k, 2);
//     final int mod5 = lucasTheorem(n, k, 5);
//     int[][] lookup = {
//         {0, 6, 2, 8, 4}, // mod2 == 0
//         {5, 1, 7, 3, 9}  // mod2 == 1
//     };
//     return lookup[mod2][mod5];
//   }
// 
//   // Returns (n, k) % prime.
//   private int lucasTheorem(int n, int k, int prime) {
//     int res = 1;
//     while (n > 0 || k > 0) {
//       final int nMod = n % prime;
//       final int MOD = k % prime;
//       res *= nCk(nMod, MOD);
//       res %= prime;
//       n /= prime;
//       k /= prime;
//     }
//     return res;
//   }
// 
//   // Returns (n, k).
//   private int nCk(int n, int k) {
//     int res = 1;
//     for (int i = 0; i < k; ++i) {
//       res *= (n - i);
//       res /= (i + 1);
//     }
//     return res;
//   }
// }

// Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3463: Check If Digits Are Equal in String After Operations II
// class Solution {
//   // Same as 3461. Check If Digits Are Equal in String After Operations I
//   public boolean hasSameDigits(String s) {
//     final int n = s.length();
//     int num1 = 0;
//     int num2 = 0;
// 
//     for (int i = 0; i + 1 < n; ++i) {
//       final int coefficient = nCMOD10(n - 2, i);
//       num1 += (coefficient * (s.charAt(i) - '0')) % 10;
//       num1 %= 10;
//       num2 += (coefficient * (s.charAt(i + 1) - '0')) % 10;
//       num2 %= 10;
//     }
// 
//     return num1 == num2;
//   }
// 
//   // Returns (n, k) % 10.
//   private int nCMOD10(int n, int k) {
//     final int mod2 = lucasTheorem(n, k, 2);
//     final int mod5 = lucasTheorem(n, k, 5);
//     int[][] lookup = {
//         {0, 6, 2, 8, 4}, // mod2 == 0
//         {5, 1, 7, 3, 9}  // mod2 == 1
//     };
//     return lookup[mod2][mod5];
//   }
// 
//   // Returns (n, k) % prime.
//   private int lucasTheorem(int n, int k, int prime) {
//     int res = 1;
//     while (n > 0 || k > 0) {
//       final int nMod = n % prime;
//       final int MOD = k % prime;
//       res *= nCk(nMod, MOD);
//       res %= prime;
//       n /= prime;
//       k /= prime;
//     }
//     return res;
//   }
// 
//   // Returns (n, k).
//   private int nCk(int n, int k) {
//     int res = 1;
//     for (int i = 0; i < k; ++i) {
//       res *= (n - i);
//       res /= (i + 1);
//     }
//     return res;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.