int EvaluateExpression(char* expr) {
    std::stack<int> numbers;
    std::stack<char> operators;

    int i = 0;
    while (expr[i] != '\0') {
        if (expr[i] == ' ') {
            i++;
            continue;
        }
        if (expr[i] >= '0' && expr[i] <= '9') {
            int num = 0;
            while (expr[i] >= '0' && expr[i] <= '9') {
                num = num * 10 + (expr[i] - '0');
                i++;
            }
            numbers.push(num);
        } else if (expr[i] == '+' expr[i] == '/') {
            while (!operators.empty() && (operators.top() == '/' operators.top() == '+')) {
                int num2 = numbers.top();
                numbers.pop();
                int num1 = numbers.top();
                numbers.pop();
                char op = operators.top();
                operators.pop();
                if (op == '/') {
                    numbers.push(num1 / num2);
                } else if (op == '+') {
                    numbers.push(num1 + num2);
                }
            }
            operators.push(expr[i]);
            i++;
        }
    }

    while (!operators.empty()) {
        int num2 = numbers.top();
        numbers.pop();
        int num1 = numbers.top();
        numbers.pop();
        char op = operators.top();
        operators.pop();
        if (op == '/') {
            numbers.push(num1 / num2);
        } else if (op == '+') {
            numbers.push(num1 + num2);
        }
    }

    return numbers.top();
}


C++✅
Accenture

int LetteredNumberSum(char* str, int len) {
    if (str == nullptr) {
        return 0;
    }

    int sum = 0;
    for (int i = 0; i < len; i++) {
        char letter = str[i];
        int ans = 0;

        switch (letter) {
            case 'A':
                ans = 1;
                break;
            case 'B':
                ans = 10;
                break;
            case 'C':
                ans = 100;
                break;
            case 'D':
                ans = 1000;
                break;
            case 'E':
                ans = 10000;
                break;
            case 'F':
                ans = 100000;
                break;
            case 'G':
                ans = 1000000;
                break;
            default:
                ans = 0;
                break;
        }

        sum += ans;
    }

    return sum;
}

C++✅

int AddDistinctSubDuplicate(int a, int b, int c, int d) {
    int sumDistinct = 0;
    int duplicateNumber = 0;

    // Calculate sum of distinct numbers
    if (a != b && a != c && a != d) {
        sumDistinct += a;
    }
    if (b != a && b != c && b != d) {
        sumDistinct += b;
    }
    if (c != a && c != b && c != d) {
        sumDistinct += c;
    }
    if (d != a && d != b && d != c) {
        sumDistinct += d;
    }

    // Find the duplicate number
    if (a == b && a != c && a != d) {
        duplicateNumber = a;
    }
    if (a == c && a != b && a != d) {
        duplicateNumber = a;
    }
    if (a == d && a != b && a != c) {
        duplicateNumber = a;
    }
    if (b == c && b != a && b != d) {
        duplicateNumber = b;
    }
    if (b == d && b != a && b != c) {
        duplicateNumber = b;
    }
    if (c == d && c != a && c != b) {
        duplicateNumber = c;
    }

    int result = sumDistinct - duplicateNumber - 308;
    return result;
}

C++✅

#include <bits/stdc++.h>
using namespace std;

void solve(int k) {
   string s = to_string(k);
   int v[] = {2,3,4,5,6};
   cout<< ((s[0] - '0') - 1) * 10 + v[s.length()] << endl;
}
int main() {
   int k =4 ; N = 5;
   solve(k);
   return 0;
}

C++✅
Accenture

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