diff --git a/Algorithms/Easy/1827_Minimum Operations to Make the Array Increasing/Solution.cpp b/Algorithms/Easy/1827_Minimum Operations to Make the Array Increasing/Solution.cpp
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+++ b/Algorithms/Easy/1827_Minimum Operations to Make the Array Increasing/Solution.cpp	
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+// https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/
+
+// Test Case
+
+// Input: nums = [1,1,1]
+// Output: 3
+// Explanation: You can do the following operations:
+// 1) Increment nums[2], so nums becomes [1,1,2].
+// 2) Increment nums[1], so nums becomes [1,2,2].
+// 3) Increment nums[2], so nums becomes [1,2,3].
+
+// Approach
+/*
+The problem can be solved as -
+    1. Just iterate over the array.
+    2. If at any point, nums[i] <= nums[i - 1], then we need to increment nums[i] to make the array strictly increasing. The number of increments needed is given by - nums[i - 1] + nums[i] + 1. Basically, it is the number of increments needed to take nums[i] to atleast nums[i - 1] + 1.
+    3.Return the number of increments.
+*/
+
+// Code
+int minOperations(vector<int> &nums)
+{
+    int count = 0;
+    for (int i = 1; i < nums.size(); i++)
+        if (nums[i] <= nums[i - 1])
+        {
+            count += (nums[i - 1] - nums[i] + 1), // nums[i] must be made atleast equal to nums[i-1]+1
+                nums[i] = nums[i - 1] + 1;
+        }
+    return count;
+}
+
+// Time Complexity : O(N)
+// Space Complexity : O(1)