From e4893b7efe31cd455907c2c86c4932a0f7f42fd4 Mon Sep 17 00:00:00 2001
From: chayan das <110921638+Chayandas07@users.noreply.github.com>
Date: Thu, 3 Oct 2024 22:15:58 +0530
Subject: [PATCH] Create 1590. Make Sum Divisible by P

---
 1590. Make Sum Divisible by P | 36 +++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 1590. Make Sum Divisible by P

diff --git a/1590. Make Sum Divisible by P b/1590. Make Sum Divisible by P
new file mode 100644
index 0000000..a4bcf89
--- /dev/null
+++ b/1590. Make Sum Divisible by P	
@@ -0,0 +1,36 @@
+typedef long long ll;
+class Solution {
+public:
+    int minSubarray(vector<int>& nums, int p) {
+        /*
+          3 1 4 2 totalsum=10
+          remove 3 or 31 or 314 
+          remove 1 or 14 or 142
+          remove 4 or 42
+          remove 2
+             0 3 1 4 2
+          ps[0 3 4 8 10]
+         brute force O(N*2);
+         how to do it nlogn or n 
+         if on subtracting from sum a number which has same remainder when%p
+         thats the ans
+         now question becomes finding subarrya with sum as sum%p
+        */
+        int n=nums.size();
+        ll totalsum=0;
+        for(auto& i:nums)totalsum+=i;
+        if(totalsum%p==0)return 0;
+        ll rem=totalsum%p;
+        unordered_map<ll,int>mp;
+        ll sum=0;int ans=n;
+        mp[0]=-1;
+        for(int i=0;i<n;i++){
+            sum = (sum+nums[i])%p;
+            ll remsum=(sum-rem+p)%p;
+            if(mp.find(remsum)!=mp.end())ans=min(ans,i-mp[remsum]);
+            mp[sum]=i;
+        }
+        if(ans==n)return -1;
+        return ans;
+    }
+};