[Shreyash]: Added code to do conversion of temperature units degC and degF

Author: shreyashdj

Functions Problems/TempConversion.js

@@ -0,0 +1,48 @@
+/*
+    Help user find degF or degC based on their Conversion Selection. Use
+    Case Statement and ensure that the inputs are within the Freezing Point (
+    0 °C / 32 °F ) and the Boiling Point of Water ( 100 °C / 212 °F )
+    a. degF = (degC * 9/5) + 32
+    b. degC = (degF – 32) * 5/9
+*/
+
+/*
+    Function to convert degC to degF
+*/
+let degC_To_DegF = function (degC) {
+    return degF = (degC * 9 / 5) + 32;
+}
+
+/*
+    Function to convert degF to degC
+*/
+let degF_to_DegC = function (degF) {
+    return degC = (degF - 32) * (5 / 9);
+}
+
+/*
+    Main code for temperature conversion
+*/
+const prompt = require('prompt-sync')();
+console.log();
+let selection = parseInt(prompt(" Enter what your want to do =>   1) degC to degF , 2) degF to degC  => "));
+switch (selection) {
+    case 1:
+        let degC = prompt(" Enter temprature -> ");
+        if (0 <= degC && degC <= 100) {
+            console.log(" " + degC + " degC = " + degC_To_DegF(degC) + " degF \n");
+        } else {
+            console.log(" Invalid Temperature \n")
+        }
+        break;
+    case 2:
+        let degF = prompt(" Enter temprature -> ");
+        if (32 <= degF && degF <= 212) {
+            console.log(" " + degF + " degF = " + degF_to_DegC(degF) + " degC \n");
+        } else {
+            console.log(" Invalid Temperature \n")
+        }
+        break;
+    default:
+        console.log(" Invalid Input \n")
+}