From 36029049668730873f6f9933828a51dab33742a9 Mon Sep 17 00:00:00 2001
From: sahil-13399 <46062089+sahil-13399@users.noreply.github.com>
Date: Mon, 4 Oct 2021 23:32:18 +0530
Subject: [PATCH] Implement MinMax solution using Stack (#2482)

Co-authored-by: sahil.samantaray <sahil.samantaray@nymble.in>
---
 .../Stacks/MaximumMinimumWindow.java          | 95 +++++++++++++++++++
 1 file changed, 95 insertions(+)
 create mode 100644 DataStructures/Stacks/MaximumMinimumWindow.java

diff --git a/DataStructures/Stacks/MaximumMinimumWindow.java b/DataStructures/Stacks/MaximumMinimumWindow.java
new file mode 100644
index 00000000..e321e65d
--- /dev/null
+++ b/DataStructures/Stacks/MaximumMinimumWindow.java
@@ -0,0 +1,95 @@
+package DataStructures.Stacks;
+
+import java.util.Arrays;
+import java.util.Stack;
+
+/**
+ * Given an integer array. The task is to find the maximum of the minimum of every window size in the array.
+ * Note: Window size varies from 1 to the size of the Array.
+ * <p>
+ * For example,
+ * <p>
+ * N = 7
+ * arr[] = {10,20,30,50,10,70,30}
+ * <p>
+ * So the answer for the above would be : 70 30 20 10 10 10 10
+ * <p>
+ * We need to consider window sizes from 1 to length of array in each iteration.
+ * So in the iteration 1 the windows would be [10], [20], [30], [50], [10], [70], [30].
+ * Now we need to check the minimum value in each window. Since the window size is 1 here the minimum element would be the number itself.
+ * Now the maximum out of these is the result in iteration 1.
+ * In the second iteration we need to consider window size 2, so there would be [10,20], [20,30], [30,50], [50,10], [10,70], [70,30].
+ * Now the minimum of each window size would be [10,20,30,10,10] and the maximum out of these is 30.
+ * Similarly we solve for other window sizes.
+ *
+ * @author sahil
+ */
+public class MaximumMinimumWindow {
+
+    /**
+     * This function contains the logic of finding maximum of minimum for every window size
+     * using Stack Data Structure.
+     *
+     * @param arr Array containing the numbers
+     * @param n Length of the array
+     * @return result array
+     */
+    public static int[] calculateMaxOfMin(int[] arr, int n) {
+        Stack<Integer> s = new Stack<>();
+        int left[] = new int[n + 1];
+        int right[] = new int[n + 1];
+        for (int i = 0; i < n; i++) {
+            left[i] = -1;
+            right[i] = n;
+        }
+
+        for (int i = 0; i < n; i++) {
+            while (!s.empty() && arr[s.peek()] >= arr[i])
+                s.pop();
+
+            if (!s.empty())
+                left[i] = s.peek();
+
+            s.push(i);
+        }
+
+        while (!s.empty())
+            s.pop();
+
+        for (int i = n - 1; i >= 0; i--) {
+            while (!s.empty() && arr[s.peek()] >= arr[i])
+                s.pop();
+
+            if (!s.empty())
+                right[i] = s.peek();
+
+            s.push(i);
+        }
+
+        int ans[] = new int[n + 1];
+        for (int i = 0; i <= n; i++)
+            ans[i] = 0;
+
+        for (int i = 0; i < n; i++) {
+            int len = right[i] - left[i] - 1;
+
+            ans[len] = Math.max(ans[len], arr[i]);
+        }
+
+        for (int i = n - 1; i >= 1; i--)
+            ans[i] = Math.max(ans[i], ans[i + 1]);
+
+        // Print the result
+        for (int i = 1; i <= n; i++)
+            System.out.print(ans[i] + " ");
+        return ans;
+    }
+
+    public static void main(String args[]) {
+        int[] arr = new int[]{10, 20, 30, 50, 10, 70, 30};
+        int[] target = new int[]{70, 30, 20, 10, 10, 10, 10};
+        int[] res = calculateMaxOfMin(arr, arr.length);
+        assert Arrays.equals(target, res);
+    }
+
+}