Check If Digits Are Equal in String After Operations II

Check If Digits Are Equal in String After Operations II - Problem

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.

Input & Output

Example 1 — Basic Reduction

$ Input: s = "1234"

› Output: false

💡 Note: Step 1: "1234" → "357" (1+2=3, 2+3=5, 3+4=7). Step 2: "357" → "82" (3+5=8, 5+7=12→2). Final: 8 ≠ 2, so return false.

Example 2 — Equal Final Digits

$ Input: s = "808"

› Output: true

💡 Note: Step 1: "808" → "88" (8+0=8, 0+8=8). Final: 8 = 8, so return true.

Example 3 — Already Two Digits

$ Input: s = "11"

› Output: true

💡 Note: String already has 2 digits: 1 = 1, so return true.

Constraints

2 ≤ s.length ≤ 1000
s consists of digits only

Visualization

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Asked in

G Google 25 M Microsoft 18

The key insight is that this reduction process follows Pascal's triangle coefficients. We can either simulate the reduction step-by-step or use mathematical properties. The simulation approach is simpler with Time: O(n²), Space: O(n), while the mathematical approach can achieve O(n) time with O(1) space.

Common Approaches

✓ Memoization

⏱️ Time: N/A Space: N/A

Pascal's Triangle Optimization

⏱️ Time: O(n) Space: O(1)

The reduction process follows Pascal's triangle coefficients. Each original digit contributes to the final result with a specific weight that can be calculated using binomial coefficients modulo 10.

Simulation Approach

⏱️ Time: O(n²) Space: O(n)

Build each new level by summing adjacent digits modulo 10, continuing until only two digits are left. Then check if they are equal.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* s) {
    int len = strlen(s);
    
    while (len > 2) {
        char newS[1000];
        int newLen = 0;
        
        for (int i = 0; i < len - 1; i++) {
            int digit1 = s[i] - '0';
            int digit2 = s[i + 1] - '0';
            int newDigit = (digit1 + digit2) % 10;
            newS[newLen++] = '0' + newDigit;
        }
        
        newS[newLen] = '\0';
        strcpy(s, newS);
        len = newLen;
    }
    
    return s[0] == s[1];
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    
    // Remove newline if present
    int len = strlen(s);
    if (len > 0 && s[len-1] == '\n') {
        s[len-1] = '\0';
    }
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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Smart Actions

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