Implement solution for "Nearest Smaller Values" problem

Author: amansharma264

Problems/Mathematics/Day-05/sol/solution1.cpp

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+/*
+Problem Name: Nearest Smaller Values
+
+Short Problem Statement:
+Given an array of n integers, for each position i, find the nearest position
+to the left that contains a strictly smaller value.
+If no such position exists, print 0 for that index.
+
+Approach (Using Prefix Sums / Stack Concept):
+We process the array from left to right using a monotonic increasing stack.
+Each stack element stores:
+- the value
+- its 1-based index
+
+For every element:
+1. Pop elements from the stack while they are greater than or equal to
+   the current value (they cannot be the nearest smaller).
+2. If the stack becomes empty, no smaller element exists on the left → print 0.
+3. Otherwise, the top of the stack gives the nearest smaller value’s position.
+4. Push the current element with its index into the stack.
+
+This ensures each element is pushed and popped at most once.
+
+Time Complexity:
+O(n)
+
+Space Complexity:
+O(n)
+
+Example:
+Input:
+8
+2 5 1 4 8 3 2 5
+
+Output:
+0 1 0 3 4 3 3 7
+
+Submission Link:
+https://cses.fi/problemset/result/15785248/
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio() ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr)
+#define ll long long
+#define endl '\n'
+
+int main() {
+    fastio();
+
+    ll n;
+    cin >> n;
+
+    vector<ll> a(n);
+    for (ll i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+
+    stack<pair<ll, ll>> st;
+
+    for (ll i = 0; i < n; i++) {
+        while (!st.empty() && st.top().first >= a[i]) {
+            st.pop();
+        }
+
+        if (st.empty()) {
+            cout << 0 << " ";
+        } else {
+            cout << st.top().second << " ";
+        }
+
+        st.push({a[i], i + 1});
+    }
+
+    return 0;
+}