No chance
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport java.util.*;\n\nclass Solution {\n    // len(subArr) >= 2\n\n    // O(N) optimal solution\n    // score[0...i] = score[0...i - 1] + 1? if 0...i - 1 DOES NOT CONTAIN\n    // (A[i] - 1) or (A[i] + 1);\n    // ALTERNATIVE VIEW: [1, 3,4] = 1 gap   [1, 3, 5] = 2 gap\n    // count GAPS between 2 consecutive elems X,Y there is a gap if Y - X > 1\n    // Problem:  for each subArrays count the number of gaps\n    // aka the number of elems A[i] where A[i + 1] - A[i] > 1\n\n    // Complication: after sorting the order will change and subArrays will be different!!\n    // We need to find out how many elems will still be part of the same subArrays\n    // after sorting and only work on those to count the final result!!!\n\n    // IDEA: Find each gap; find how many subArrays each gap adds +1 to\n    // GAP EXISTS when:\n    // subArray contains A[i] and NOT (A[i] + 1) but some (A[i] + X)\n    //\n\n    // 3,4,1   score = 2\n    // [1,3,4] [1,4]    invalid: [3,4]     [1] [3] [4]  ALL SUBARRAYS\n    // {1,3} gap => [1,3,4]\n    // {1,4} gap => [1,4]\n\n    // A[i] will form gap with any (A[i] + X) where X > 1, on LEFT and RIGHT\n    // we can keep expanding left and right adding elems until we reach INVALID\n    // boundary (A[i] + 1)\n    // we pretend A[i] is center of subarray and build (numLeft * numRight)\n    // subArray combinations, each adding +1 to final result!!!!\n    // in the combinations LEFT * RIGHT example:\n    // [1,3,3,3,5] so:\n    // [{1,3},3,3,5] =>  [1,3] [1,3,3] [1,3,3,3] [1,3,3,3,5]\n    // [1,3,3,{3,5}] => 4 (left) * 1 (right) ..\n\n    // BOUNDARIES for A[i]:\n    // 1) A[i] if ON RIGHT SIDE: because will not be moved after sorting!!!\n    // all A[i] duplicates are valid additions on the left\n    // 2) A[i + 1]\n\n    // NEED (A[i] + X) to make a valid gap!!!!! pair with closest left/right BIGGER\n    // But we never know if A[i] is the biggest of the curr subArray, so:\n    // Instead just subtract from the final result the number of times we\n    // overcounted gaps using A[i] where A[i] was the biggest elem of the subArray\n    // But each subarray has 1 max val; so this = number of all subArrays\n    // N * (N + 1) / 2\n    public int sumImbalanceNumbers(int[] A) {\n        int n = A.length;\n        int cnt = 0;\n\n        int[] rightBoundary = new int[n + 2]; // first (A[i] + 1) or A[i]\n        int[] valToLastId = new int[n + 2];\n        Arrays.fill(valToLastId, n);\n        Arrays.fill(rightBoundary, n);\n        for (int i = n - 1; i >= 0; i--) {\n            rightBoundary[i] = Math.min(valToLastId[A[i]], valToLastId[A[i] + 1]);\n            valToLastId[A[i]] = i;\n        }\n        Arrays.fill(valToLastId, -1); // reuse as left boundary, last (A[i] + 1)\n        for (int i = 0; i < n; i++) {\n            int numLeft = i - valToLastId[A[i] + 1];\n            int numRight = rightBoundary[i] - i;\n            cnt += numLeft * numRight;\n            valToLastId[A[i]] = i;\n        }\n\n        int numOvercounted = n * (n + 1) / 2; // handles cases of subarrays [X] singletons\n        return cnt - numOvercounted;\n    } // 1 + 4 + 4 + 4        n = 4       -10\n\n\n    // score[0...i] = score[0...i - 1] + 1? if 0...i - 1 DOES NOT CONTAIN\n    // (A[i] - 1) or (A[i] + 1); if it does, then adding this elem gives no +1 score\n    // because it effectively merges with an old elem;\n    // if it contains BOTH (A[i] - 1) AND (A[i] + 1) and NOT A[i] then adding it\n    // bridges 2 islands and score -1\n    // score(3,5) = 1  add 4 => score(3,4,5) = 0\n//    // O(N^2) brute-force\n//    public int sumImbalanceNumbers(int[] A) {\n//        int n = A.length;\n//        int cnt = 0;\n//        boolean[] added = null;\n//        for (int i = 0; i < n; i++) {\n//            added = new boolean[n + 2];\n//            int score = 0;\n//            for (int j = i; j < n; j++) {\n//                int curr = A[j];\n//                if (!added[curr]) { // elem chages; update score of curr subArr\n//                    added[curr] = true;\n//                    if (j == i) continue;\n//\n//                    int next = curr + 1;\n//                    int prev = curr - 1;\n//                    if (added[next] && added[prev]) {\n//                        score--;\n//                    } else if (!added[next] && !added[prev]) {\n//                        score++;\n//                    }\n//                }\n//                score = Math.max(0, score);\n//                cnt += score;\n//            }\n//        }\n//\n//        return cnt;\n//    }\n\n\n}\n\n```
['Java']