Sum of all possible expressions of a numeric string possible by inserting addition operators
Given a numeric string str of length N, the task is to find the sum of all possible expressions by inserting the ‘+’ operator between the characters of the string any number of times.
Examples:
Input: str = “125” Output: 176 Explanation: Inserting “+” after 1st index modifies str to “1+25” and value = 26 Inserting “+” after 2nd index modifies str to “12+5” and value = 17 Inserting “+” after both 1st and 2nd index modifies str to “1+2+5” and value = 8 Therefore, the total sum of all possible expression is 125 + 26 + 17 + 8 = 176
Input: str = “9999999999”
Output: 12656242944
Approach: The idea is to insert the ‘+’ operator at all possible index of the string in all possible ways and calculate the sum. Finally, print the total sum obtained. Follow the steps below to solve the problem:
- Initialize a variable say, sumOfExp to store the sum of all possible expression by inserting the ‘+’ operator at all possible indices of the string.
- Generate all possible subset of indices of the string iteratively. For every subset of indices inserts the ‘+’ operator at elements of the subset and increment sumOfExp by the sum of the current expression.
- Finally, print the value of sumOfExp.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find sum of all expressions by// inserting '+' operator at all possible indicesvoid findSumOfExpressions(string S, int N){ // Stores sum of all expressions by inserting // '+' operator at all possible indices unsigned long long sumOfExp = 0; // Generate all possible subset // of indices iteratively for (int i = 0; i < pow(2, N - 1); i++) { // Stores sum of // current expressions unsigned long long ans_sub = 0; // Stores numbers of // current expressions string subst = string(1, S.at(0)); // Traverse the string at insert + at // current subset of indices for (int j = 0; j < N - 1; j++) { // If current index exists // in the current subset if (((i >> j) & 1) == 1) { // Update ans_sub ans_sub += stoull(subst); // Update subset subst = string(1, S.at(j + 1)); } else // Update subset subst += S.at(j + 1); // + can't be inserted after // the last index if (j == N - 2) ans_sub += stoull(subst); } // Update ans sumOfExp += ans_sub; } // Base case if (N == 1) cout << S; else // Print answer cout << sumOfExp;}// Driver Codeint main(){ // Given string string S = "9999999999"; // Length of the string int N = S.length(); // Function call findSumOfExpressions(S, N);}// This code is contributed by phasing17. |
Java
Python3
C#
Javascript
12656242944
Time Complexity: O(2N * N)
Auxiliary Space: O(1)
