Find minimum value expression by inserting addition or multiplication operator between digits of given number
Last Updated :
26 Oct, 2021
Given string str that contains only digits, the task is to return an expression by inserting the ‘+’ or ‘*’ operator between every two digits such that the arithmetic value of the expression is minimized.
Example:
Input: str = “322”
Output: “3+2*2”
Explanation: The value of above expression is 7 which is minimum possible over the other expressions “3+2+2”, “3*2+2”, “3*3*2”
Input: str = “391118571”
Output: “3+9*1*1*1+8+5+7*1”
Approach: The given problem can be solved using a greedy approach. Follow the steps below to solve the problem:
- If the string str contains character ‘0’ then:
- Insert multiplication operator between every character of the string
- Else create a string ans to store the iterate the string and if the current character str[i] is:
- ‘1’, then Insert multiplication operator ‘*’ between the current and previous character
- not ‘1’ then Insert addition operator ‘+’ between the current and previous character
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string minimalExpression(string str)
{
string ans = "";
ans.push_back(str[0]);
bool iszero = false;
for (int i = 0; i < str.size();
i++) {
if (str[i] == '0') {
iszero = true;
}
}
if (iszero) {
for (int i = 1; i < str.size();
i++) {
ans.push_back('*');
ans.push_back(str[i]);
}
return ans;
}
for (int i = 1; i < str.size(); i++) {
if (str[i] == '1') {
ans.push_back('*');
ans.push_back(str[i]);
}
else {
ans.push_back('+');
ans.push_back(str[i]);
}
}
return ans;
}
int main()
{
string str = "391118571";
cout << minimalExpression(str);
}
|
Java
class GFG {
public static String minimalExpression(String str) {
String ans = "";
ans = ans + str.charAt(0);
boolean iszero = false;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '0') {
iszero = true;
}
}
if (iszero) {
for (int i = 1; i < str.length(); i++) {
ans += '*';
ans += str.charAt(i);
}
return ans;
}
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i) == '1') {
ans += '*';
ans += str.charAt(i);
}
else {
ans += '+';
ans += str.charAt(i);
}
}
return ans;
}
public static void main(String args[]) {
String str = "391118571";
System.out.println(minimalExpression(str));
}
}
|
Python3
def minimalExpression(str):
ans = ""
ans += str[0]
iszero = False
for i in range(0, len(str)):
if (str[i] == '0'):
iszero = True
if (iszero):
for i in range(1, len(str)):
ans += '*'
ans += str[i]
return ans
for i in range(1, len(str)):
if (str[i] == '1'):
ans += '*'
ans += str[i]
else:
ans += '+'
ans += str[i]
return ans
if __name__ == "__main__":
str = "391118571"
print(minimalExpression(str))
|
C#
using System;
class GFG {
static string minimalExpression(string str)
{
string ans = "";
ans += str[0];
bool iszero = false;
for (int i = 0; i < str.Length; i++) {
if (str[i] == '0') {
iszero = true;
}
}
if (iszero) {
for (int i = 1; i < str.Length; i++) {
ans += '*';
ans += (str[i]);
}
return ans;
}
for (int i = 1; i < str.Length; i++) {
if (str[i] == '1') {
ans += '*';
ans += (str[i]);
}
else {
ans += '+';
ans += (str[i]);
}
}
return ans;
}
public static void Main()
{
string str = "391118571";
Console.WriteLine(minimalExpression(str));
}
}
|
Javascript
<script>
function minimalExpression(str) {
let ans = [];
ans.push(str[0]);
let iszero = false;
for (let i = 0; i < str.length; i++) {
if (str[i] == "0") {
iszero = true;
}
}
if (iszero) {
for (let i = 1; i < str.length; i++) {
ans.push("*");
ans.push(str[i]);
}
return ans;
}
for (let i = 1; i < str.length; i++) {
if (str[i] == "1") {
ans.push("*");
ans.push(str[i]);
}
else {
ans.push("+");
ans.push(str[i]);
}
}
return ans.join("");
}
let str = "391118571";
document.write(minimalExpression(str));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the length of the given string
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