From d3415eb4ede0c67098433eda84e84843267a7d62 Mon Sep 17 00:00:00 2001
From: Paulo Custodio <pauloscustodio@gmail.com>
Date: Tue, 24 Sep 2024 10:23:09 +0100
Subject: Add Python solution to challenge 074

---
 challenge-074/paulo-custodio/python/ch-1.py | 45 ++++++++++++++++++++++++
 challenge-074/paulo-custodio/python/ch-2.py | 54 +++++++++++++++++++++++++++++
 2 files changed, 99 insertions(+)
 create mode 100644 challenge-074/paulo-custodio/python/ch-1.py
 create mode 100644 challenge-074/paulo-custodio/python/ch-2.py

(limited to 'challenge-074/paulo-custodio/python')

diff --git a/challenge-074/paulo-custodio/python/ch-1.py b/challenge-074/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..cce27f69b9
--- /dev/null
+++ b/challenge-074/paulo-custodio/python/ch-1.py
@@ -0,0 +1,45 @@
+#!/usr/bin/env python3
+
+# Challenge 074
+#
+# TASK #1 > Majority Element
+# Submitted by: Mohammad S Anwar
+# You are given an array of integers of size $N.
+#
+# Write a script to find the majority element. If none found then print -1.
+#
+# Majority element in the list is the one that appears more than
+# floor(size_of_list/2).
+#
+# Example 1
+# Input: @A = (1, 2, 2, 3, 2, 4, 2)
+# Output: 2, as 2 appears 4 times in the list which is more than floor(7/2).
+#
+# Example 2
+# Input: @A = (1, 3, 1, 2, 4, 5)
+# Output: -1 as none of the elements appears more than floor(6/2).
+
+import sys
+
+def majority_elem(a):
+    # count instances of each element, get max
+    count = {}
+    max_count = 0
+    max_elem = None
+    for x in a:
+        if x in count:
+            count[x] += 1
+        else:
+            count[x] = 1
+
+        if count[x] > max_count:
+            max_count, max_elem = count[x], x
+
+    # check if majority
+    if max_count > int(len(a)/2):
+        return max_elem
+    else:
+        return -1
+
+a = [int(x) for x in sys.argv[1:]]
+print(majority_elem(a))
diff --git a/challenge-074/paulo-custodio/python/ch-2.py b/challenge-074/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..131c393c0e
--- /dev/null
+++ b/challenge-074/paulo-custodio/python/ch-2.py
@@ -0,0 +1,54 @@
+#!/usr/bin/env python3
+
+# Challenge 074
+#
+# TASK #2 > FNR Character
+# Submitted by: Mohammad S Anwar
+# You are given a string $S.
+#
+# Write a script to print the series of first non-repeating character (left
+# -> right) for the given string. Print # if none found.
+#
+# Example 1
+# Input: $S = 'ababc'
+# Output: 'abb#c'
+# Pass 1: "a", the FNR character is 'a'
+# Pass 2: "ab", the FNR character is 'b'
+# Pass 3: "aba", the FNR character is 'b'
+# Pass 4: "abab", no FNR found, hence '#'
+# Pass 5: "ababc" the FNR character is 'c'
+#
+# Example 2
+# Input: $S = 'xyzzyx'
+# Output: 'xyzyx#'
+# Pass 1: "x", the FNR character is "x"
+# Pass 2: "xy", the FNR character is "y"
+# Pass 3: "xyz", the FNR character is "z"
+# Pass 4: "xyzz", the FNR character is "y"
+# Pass 5: "xyzzy", the FNR character is "x"
+# Pass 6: "xyzzyx", no FNR found, hence '#'
+
+import sys
+
+def lnr_char(s):
+    out = ""
+    seen = {}
+    for i in range(len(s)):
+        ch = s[i]
+        if ch in seen:
+            seen[ch] += 1
+        else:
+            seen[ch] = 1
+        found = False
+        for j in range(i+1)[::-1]:
+            ch = s[j]
+            if ch in seen and seen[ch] == 1:
+                out += ch
+                found = True
+                break
+        if not found:
+            out += "#"
+    return out
+
+s = sys.argv[1]
+print(lnr_char(s))
-- 
cgit