Add D1P1 in awk

2019/04/day4.awk

@@ -0,0 +1,26 @@
+#!/usr/bin/awk -f
+
+function isAscending(number) {
+    digit1 = int(substr(number, 1, 1));
+    digit2 = int(substr(number, 2, 1));
+    digit3 = int(substr(number, 3, 1));
+    digit4 = int(substr(number, 4, 1));
+    digit5 = int(substr(number, 5, 1));
+    digit6 = int(substr(number, 6, 1));
+    return (digit1 <= digit2 && digit2 <= digit3 && digit3 <= digit4 && digit4 <= digit5 && digit5 <= digit6);
+}
+
+{
+    split($1, inputs, "-");
+    lower = inputs[1];
+    upper = inputs[2];
+    for (i=lower; i<upper; i++) {
+        iString = i + "";
+        # this is what I would have done, but awk regex doesn’t support back references
+        #if (match(iString, "(.)\1") != 0 && isAscending(iString)) {
+        if (isAscending(iString))
+            print iString;
+            #valid++;
+    }
+}
+

2019/04/input

@@ -0,0 +1 @@
+172930-683082

2019/04/run_day4.sh

@@ -0,0 +1,3 @@
+#!/bin/sh
+
+awk -f day4.awk input | rg --pcre2 '(.)\1' | wc -l | awk '{printf("Part 1: %s\n", $1);}'