# Integer to English - Amazon Top Interview Questions

### Problem Statement :

Given a non-negative integer num, convert it to an English number word as a string. num will be less than one trillion.

Constraints

0 ≤ num < 10 ** 12

Example 1

Input

num = 523

Output

"Five Hundred Twenty Three"

Example 2

Input

num = 823418

Output

"Eight Hundred Twenty Three Thousand Four Hundred Eighteen"

Example 3

Input

num = 700

Output

"Seven Hundred"

### Solution :

Solution in C++ :

vector<pair<string, int>> tbl{{"Billion", 1000000000},
{"Million", 1000000},
{"Thousand", 1000},
{"Hundred", 100},
{"Ninety", 90},
{"Eighty", 80},
{"Seventy", 70},
{"Sixty", 60},
{"Fifty", 50},
{"Forty", 40},
{"Thirty", 30},
{"Twenty", 20},
{"Nineteen", 19},
{"Eighteen", 18},
{"Seventeen", 17},
{"Sixteen", 16},
{"Fifteen", 15},
{"Fourteen", 14},
{"Thirteen", 13},
{"Twelve", 12},
{"Eleven", 11},
{"Ten", 10},
{"Nine", 9},
{"Eight", 8},
{"Seven", 7},
{"Six", 6},
{"Five", 5},
{"Four", 4},
{"Three", 3},
{"Two", 2},
{"One", 1}};
string solve(int num) {
if (num == 0) return "Zero";
string ans;
for (auto& p : tbl) {
if (p.second <= num) {
if (p.second >= 100) {
ans = solve(num / p.second) + " " + p.first;
if (num > (num / p.second) * p.second)
ans += " " + solve(num - (num / p.second) * p.second);
} else {
ans = p.first + (num > p.second ? " " + solve(num - p.second) : "");
}
break;
}
}
return ans;
}

Solution in Java :

import java.util.*;

class Solution {
public String solve(int num) {
if (num == 0) {
return "Zero";
}
StringBuilder sb = new StringBuilder();
if (num >= 1000000000) {
num %= 1000000000;
}
if (num >= 1000000) {
num %= 1000000;
}
if (num >= 1000) {
num %= 1000;
}
}

private StringBuilder oneToNineHundreadNinetyNine(StringBuilder sb, int num) {
if (num >= 100) {
sb.append(oneToNineteen(num / 100)).append("Hundred ");
num %= 100;
}
if (num >= 20) {
sb.append(twentyToNinety(num));
num %= 10;
}
sb.append(oneToNineteen(num));
return sb;
}

private String oneToNineteen(int num) {
switch (num % 20) {
default:
return "";
case 1:
return "One ";
case 2:
return "Two ";
case 3:
return "Three ";
case 4:
return "Four ";
case 5:
return "Five ";
case 6:
return "Six ";
case 7:
return "Seven ";
case 8:
return "Eight ";
case 9:
return "Nine ";
case 10:
return "Ten ";
case 11:
return "Eleven ";
case 12:
return "Twelve ";
case 13:
return "Thirteen ";
case 14:
return "Fourteen ";
case 15:
return "Fifteen ";
case 16:
return "Sixteen ";
case 17:
return "Seventeen ";
case 18:
return "Eighteen ";
case 19:
return "Nineteen ";
}
}

private String twentyToNinety(int num) {
switch (num / 10 % 10) {
default:
return "";
case 2:
return "Twenty ";
case 3:
return "Thirty ";
case 4:
return "Forty ";
case 5:
return "Fifty ";
case 6:
return "Sixty ";
case 7:
return "Seventy ";
case 8:
return "Eighty ";
case 9:
return "Ninety ";
}
}
}

Solution in Python :

class Solution:
def solve(self, num):
if not num:
return "Zero"
_0_to_19 = [
"Zero",
"One",
"Two",
"Three",
"Four",
"Five",
"Six",
"Seven",
"Eight",
"Nine",
"Ten",
"Eleven",
"Twelve",
"Thirteen",
"Fourteen",
"Fifteen",
"Sixteen",
"Seventeen",
"Eighteen",
"Nineteen",
]
_0_to_90_by_10 = [
"",
"",
"Twenty",
"Thirty",
"Forty",
"Fifty",
"Sixty",
"Seventy",
"Eighty",
"Ninety",
]
_0_to_tri_by_1000 = [
"",
"Thousand",
"Million",
"Billion",
"Trillion",
]

def say_19(x):
return x and [_0_to_19[x]] or []

def say_99(x):
return x >= 20 and [_0_to_90_by_10[x // 10]] + say_19(x % 10) or say_19(x)

def say_999(x, suf):
return (
(x >= 100 and [_0_to_19[x // 100]] + ["Hundred"] or [])
+ say_99(x % 100)
+ (x and suf and [suf] or [])
)

def say(x, p):
return x and say(x // 1000, p + 1) + say_999(x % 1000, _0_to_tri_by_1000[p]) or []

return " ".join(say(num, 0))

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